GSEB Class 11 Maths Solutions Chapter 13 Limits and Derivatives Exercise 13.1

Question 1. Evaluate the following limits: \( \lim_{x \rightarrow 3} (x + 3) \)
Answer: To find the limit, we directly substitute the value \( x=3 \) into the expression. This gives us \( 3+3 = 6 \).
In simple words: When \( x \) gets very close to 3, the value of \( x+3 \) gets very close to 6.

Exam Tip: For polynomial functions, you can often find the limit by simply substituting the value \( x \) approaches directly into the expression.

Question 2. Evaluate the following limits: \( \lim_{x \rightarrow \pi} (x - \frac{22}{7}) \)
Answer: To find the limit, we just substitute \( x=\pi \) into the expression. So, the limit is \( \pi - \frac{22}{7} \).
In simple words: As \( x \) gets very close to \( \pi \), the value of \( x - \frac{22}{7} \) approaches \( \pi - \frac{22}{7} \).

Exam Tip: Remember that \( \pi \) is a constant number, even though its decimal representation is non-repeating and non-terminating. Treat it like any other number when substituting into limits.

Question 3. Evaluate the following limits: \( \lim_{r \rightarrow 1} (\pi r^2) \)
Answer: To find this limit, we substitute \( r=1 \) into the expression. This gives us \( \pi (1)^2 = \pi \cdot 1 = \pi \).
In simple words: When \( r \) gets very close to 1, the value of \( \pi r^2 \) approaches \( \pi \).

Exam Tip: Constants like \( \pi \) remain unchanged during the limit evaluation process, only variables are substituted or simplified.

Question 4. Evaluate the following limits: \( \lim_{x \rightarrow 4} \frac{4x+3}{x-2} \)
Answer: Since the denominator is not zero when \( x=4 \), we can directly substitute \( x=4 \) into the expression. This yields: \[ \frac{4(4)+3}{4-2} = \frac{16+3}{2} = \frac{19}{2} \]
In simple words: We can put 4 in place of \( x \) because the bottom part won't become zero. Doing so gives us \( \frac{19}{2} \).

Exam Tip: Always check if direct substitution results in an indeterminate form (like \( \frac{0}{0} \)) before proceeding. If not, direct substitution is the simplest method.

Question 5. Evaluate the following limits: \( \lim_{x \rightarrow -1} \frac{x^{10}+x^5+1}{x-1} \)
Answer: We substitute \( x = -1 \) into the expression. The numerator becomes \( (-1)^{10} + (-1)^5 + 1 = 1 - 1 + 1 = 1 \). The denominator becomes \( -1 - 1 = -2 \). So the limit is \( \frac{1}{-2} = -\frac{1}{2} \).
In simple words: When we put \( -1 \) into the expression, the top part becomes 1 and the bottom part becomes -2. So the answer is \( -\frac{1}{2} \).

Exam Tip: Be careful with signs when substituting negative numbers into powers. An even power makes a negative number positive, while an odd power keeps it negative.

Question 6. Evaluate the following limits: \( \lim_{x \rightarrow 0} \frac{(x+1)^5-1}{x} \)
Answer: If we try to substitute \( x=0 \) directly, we get \( \frac{0}{0} \), an indeterminate form. We need to simplify the expression first. We can expand \( (x+1)^5 \) using the binomial theorem, but a simpler way is to notice that as \( x \rightarrow 0 \), we can use the identity: \( \lim_{a \rightarrow 0} \frac{(1+a)^n-1}{a} = n \). In our case, \( a=x \) and \( n=5 \).
\( \lim_{x \rightarrow 0} \frac{(x+1)^5-1}{x} \)
\( = \lim_{x \rightarrow 0} \frac{(1 + 5x + 10x^2 + 10x^3 + 5x^4 + x^5) - 1}{x} \)
\( = \lim_{x \rightarrow 0} \frac{5x + 10x^2 + 10x^3 + 5x^4 + x^5}{x} \)
\( = \lim_{x \rightarrow 0} (5 + 10x + 10x^2 + 5x^3 + x^4) \)
Now, substitute \( x=0 \): \( 5 + 0 + 0 + 0 + 0 = 5 \).
In simple words: This problem is of the \( \frac{0}{0} \) type. We can solve it by expanding the top part or using a special formula. When we simplify and then put \( x=0 \), the answer is 5.

Exam Tip: Recognize common limit forms like \( \lim_{x \rightarrow 0} \frac{(1+x)^n-1}{x} = n \) or \( \lim_{x \rightarrow a} \frac{x^n-a^n}{x-a} = na^{n-1} \) to quickly solve such problems without full expansion.

Question 7. Evaluate the following limits: \( \lim_{x \rightarrow 2} \frac{3x^2-x-10}{x^2-4} \)
Answer: If we substitute \( x=2 \) directly, both the numerator and denominator become 0 \( (\frac{0}{0}) \). This means we need to factorize them. The numerator \( 3x^2-x-10 \) can be factored as \( (3x+5)(x-2) \). The denominator \( x^2-4 \) is a difference of squares, \( (x-2)(x+2) \).
So, the expression becomes: \[ \lim_{x \rightarrow 2} \frac{(3x+5)(x-2)}{(x-2)(x+2)} \] Since \( x \rightarrow 2 \), \( x \ne 2 \), so we can cancel out the \( (x-2) \) terms. \[ = \lim_{x \rightarrow 2} \frac{3x+5}{x+2} \] Now, substitute \( x=2 \): \[ = \frac{3(2)+5}{2+2} = \frac{6+5}{4} = \frac{11}{4} \]
In simple words: This problem gives \( \frac{0}{0} \) when we put in 2. So, we factorize the top and bottom parts to remove the common \( (x-2) \) term. After that, we can easily put \( x=2 \) to get the final answer.

Exam Tip: When faced with an indeterminate form \( \frac{0}{0} \) involving polynomials, always attempt to factorize the numerator and denominator to cancel out common factors that cause the zero.

Question 8. Evaluate the following limits: \( \lim_{x \rightarrow 3} \frac{x^4-81}{2x^2-5x-3} \)
Answer: Substituting \( x=3 \) gives \( \frac{3^4-81}{2(3)^2-5(3)-3} = \frac{81-81}{18-15-3} = \frac{0}{0} \), which is an indeterminate form. We must factorize. The numerator \( x^4-81 \) can be factored as \( (x^2-9)(x^2+9) = (x-3)(x+3)(x^2+9) \). The denominator \( 2x^2-5x-3 \) can be factored as \( (2x+1)(x-3) \).
So the limit expression becomes: \[ \lim_{x \rightarrow 3} \frac{(x-3)(x+3)(x^2+9)}{(x-3)(2x+1)} \] Since \( x \rightarrow 3 \), \( x \ne 3 \), so we can cancel the \( (x-3) \) terms: \[ = \lim_{x \rightarrow 3} \frac{(x+3)(x^2+9)}{2x+1} \] Now, substitute \( x=3 \): \[ = \frac{(3+3)(3^2+9)}{2(3)+1} = \frac{6 \times (9+9)}{6+1} = \frac{6 \times 18}{7} = \frac{108}{7} \]
In simple words: This limit is also \( \frac{0}{0} \) when you put in 3. We break down both the top and bottom parts into their factors. We then cancel out the \( (x-3) \) part that makes it \( \frac{0}{0} \). After that, we put \( x=3 \) back into the simplified expression to get the answer.

Exam Tip: Always simplify the expression by factoring and canceling common terms before attempting direct substitution when dealing with indeterminate forms. This is a crucial step for polynomial limits.

Question 9. Evaluate the following limits: \( \lim_{x \rightarrow 0} \frac{ax+b}{cx+1} \)
Answer: We can find the limit by directly substituting \( x=0 \) into the expression, because the denominator \( cx+1 \) will become 1 (not zero). \[ \frac{a(0)+b}{c(0)+1} = \frac{0+b}{0+1} = \frac{b}{1} = b \]
In simple words: We just replace \( x \) with 0 in the expression. The bottom part won't be zero, so the limit is simply \( b \).

Exam Tip: Direct substitution is valid when the function is well-defined at the limit point (i.e., the denominator is non-zero, or there are no undefined terms).

Question 10. Evaluate the following limits: \( \lim_{z \rightarrow 1} \frac{z^{\frac{1}{3}}-1}{z^{\frac{1}{6}}-1} \)
Answer: Substituting \( z=1 \) gives \( \frac{1^{\frac{1}{3}}-1}{1^{\frac{1}{6}}-1} = \frac{1-1}{1-1} = \frac{0}{0} \), an indeterminate form. Let \( y = z^{\frac{1}{6}} \). Then \( z^{\frac{1}{3}} = (z^{\frac{1}{6}})^2 = y^2 \). As \( z \rightarrow 1 \), \( y \rightarrow 1^{\frac{1}{6}} = 1 \). So the limit becomes: \[ \lim_{y \rightarrow 1} \frac{y^2-1}{y-1} \] We can factor the numerator as a difference of squares: \( y^2-1 = (y-1)(y+1) \). \[ = \lim_{y \rightarrow 1} \frac{(y-1)(y+1)}{y-1} \] Since \( y \rightarrow 1 \), \( y \ne 1 \), so we can cancel the \( (y-1) \) terms: \[ = \lim_{y \rightarrow 1} (y+1) \] Now, substitute \( y=1 \): \[ = 1+1 = 2 \]
In simple words: If we put 1 in, we get \( \frac{0}{0} \). To fix this, we can change \( z^{\frac{1}{6}} \) to a new letter, say \( y \). Then, we use the difference of squares rule to simplify and cancel parts. After that, we put 1 back into the new letter, and the answer is 2.

Exam Tip: When dealing with fractional powers, a substitution (like \( y = x^{1/n} \)) can often transform the expression into a simpler polynomial form, which is easier to factorize or apply standard limit identities.

Question 11. Evaluate the following limits: \( \lim_{x \rightarrow 1} \frac{ax^2+bx+c}{cx^2+bx+a} \)
Answer: Since the denominator \( cx^2+bx+a \) will not be zero upon direct substitution (unless \( a+b+c=0 \)), we can directly substitute \( x=1 \) into the expression. \[ = \frac{a(1)^2+b(1)+c}{c(1)^2+b(1)+a} \] \[ = \frac{a+b+c}{c+b+a} \] Assuming \( a+b+c \ne 0 \), the numerator and denominator are identical, so the value is 1.
In simple words: We can directly substitute \( x=1 \) into the expression. This makes the top part \( a+b+c \) and the bottom part \( c+b+a \). Since these are the same, the answer is 1.

Exam Tip: Always simplify algebraic expressions before concluding the final limit value. If the numerator and denominator are identical after substitution, the limit is 1.

Question 12. Evaluate the following limits: \( \lim_{x \rightarrow -2} \frac{\frac{1}{x}+\frac{1}{2}}{x+2} \)
Answer: If we substitute \( x=-2 \) directly, we get \( \frac{\frac{1}{-2}+\frac{1}{2}}{-2+2} = \frac{0}{0} \), an indeterminate form. We need to simplify the numerator first by finding a common denominator for the fractions: \[ \frac{1}{x}+\frac{1}{2} = \frac{2}{2x} + \frac{x}{2x} = \frac{2+x}{2x} \] Now substitute this back into the limit expression: \[ \lim_{x \rightarrow -2} \frac{\frac{2+x}{2x}}{x+2} = \lim_{x \rightarrow -2} \frac{2+x}{2x(x+2)} \] Since \( x \rightarrow -2 \), \( x \ne -2 \), so \( (x+2) \) is not zero and we can cancel out the \( (2+x) \) terms (which is the same as \( (x+2) \)): \[ = \lim_{x \rightarrow -2} \frac{1}{2x} \] Now, substitute \( x=-2 \): \[ = \frac{1}{2(-2)} = -\frac{1}{4} \]
In simple words: This problem becomes \( \frac{0}{0} \) if we put -2 in. We combine the fractions on top, then cancel out the \( (x+2) \) term that makes it zero. Finally, we put -2 into the simpler expression to get the answer.

Exam Tip: When fractions are involved in the numerator or denominator, simplify them by finding a common denominator before attempting to factor or cancel terms. This helps resolve indeterminate forms.

Question 13. Evaluate the following limits: \( \lim_{x \rightarrow 0} \frac{\sin ax}{bx} \)
Answer: This limit involves a trigonometric function. We use the standard limit identity \( \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \). To apply this, we adjust the expression: \[ \lim_{x \rightarrow 0} \frac{\sin ax}{bx} = \lim_{x \rightarrow 0} \left( \frac{\sin ax}{ax} \cdot \frac{ax}{bx} \right) \] We can separate the terms: \[ = \left( \lim_{x \rightarrow 0} \frac{\sin ax}{ax} \right) \cdot \left( \lim_{x \rightarrow 0} \frac{ax}{bx} \right) \] The first limit is 1 (as \( ax \rightarrow 0 \) when \( x \rightarrow 0 \)). The second limit simplifies to \( \frac{a}{b} \). \[ = 1 \cdot \frac{a}{b} = \frac{a}{b} \]
In simple words: To solve this, we make the expression look like \( \frac{\sin \theta}{\theta} \). We multiply and divide by \( ax \) to create this form. Once it's in that shape, we know that part becomes 1, leaving us with \( \frac{a}{b} \).

Exam Tip: For limits involving \( \sin x \) or \( \tan x \) as \( x \rightarrow 0 \), always aim to transform the expression into the form \( \frac{\sin \theta}{\theta} \) or \( \frac{\tan \theta}{\theta} \) to apply the fundamental trigonometric limit rules.

Question 14. Evaluate the following limits: \( \lim_{x \rightarrow 0} \frac{\sin ax}{\sin bx}, ab \ne 0 \)
Answer: We use the standard limit identity \( \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \). To apply this identity to both the numerator and denominator, we multiply and divide by appropriate terms: \[ \lim_{x \rightarrow 0} \frac{\sin ax}{\sin bx} = \lim_{x \rightarrow 0} \left( \frac{\sin ax}{ax} \cdot ax \cdot \frac{1}{\frac{\sin bx}{bx} \cdot bx} \right) \] Rearranging the terms: \[ = \lim_{x \rightarrow 0} \left( \frac{\sin ax}{ax} \cdot \frac{bx}{\sin bx} \cdot \frac{ax}{bx} \right) \] Now, we can evaluate each part: \[ = \left( \lim_{x \rightarrow 0} \frac{\sin ax}{ax} \right) \cdot \left( \lim_{x \rightarrow 0} \frac{bx}{\sin bx} \right) \cdot \left( \lim_{x \rightarrow 0} \frac{a}{b} \right) \] Each of the first two limits equals 1, and the third is a constant \( \frac{a}{b} \). \[ = 1 \cdot 1 \cdot \frac{a}{b} = \frac{a}{b} \]
In simple words: This limit has sine functions on both the top and bottom. We use the rule that \( \frac{\sin \theta}{\theta} \) goes to 1 when \( \theta \) goes to 0. We make the top look like \( \frac{\sin ax}{ax} \) and the bottom look like \( \frac{\sin bx}{bx} \). After simplifying, the answer becomes \( \frac{a}{b} \).

Exam Tip: When both numerator and denominator contain sine or tangent functions that tend to zero, a common strategy is to multiply and divide by the arguments of the trigonometric functions to leverage the standard limit form \( \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \).

Question 15. Evaluate the following limits: \( \lim_{x \rightarrow \pi} \frac{\sin (\pi-x)}{\pi(\pi-x)} \)
Answer: This limit is in the form \( \frac{0}{0} \) if we substitute \( x=\pi \). Let \( \theta = \pi - x \). As \( x \rightarrow \pi \), \( \theta \rightarrow \pi - \pi = 0 \). Substitute \( \theta \) into the limit expression: \[ \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\pi \theta} \] We can take the constant \( \frac{1}{\pi} \) outside the limit: \[ = \frac{1}{\pi} \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} \] We know that \( \lim_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \). \[ = \frac{1}{\pi} \cdot 1 = \frac{1}{\pi} \]
In simple words: If we put \( \pi \) in, we get \( \frac{0}{0} \). We change the variable by letting \( \theta = \pi - x \). As \( x \) goes to \( \pi \), \( \theta \) goes to 0. This makes the limit a standard form \( \frac{\sin \theta}{\theta} \), so the answer is \( \frac{1}{\pi} \).

Exam Tip: For limits involving \( (a-x) \) or \( (x-a) \) within trigonometric functions as \( x \rightarrow a \), a substitution like \( \theta = a-x \) or \( \theta = x-a \) often simplifies the problem to a standard limit form.

Question 16. Evaluate the following limits: \( \lim_{x \rightarrow 0} \frac{\cos x}{\pi-x} \)
Answer: We can directly substitute \( x=0 \) into the expression because the denominator \( \pi-x \) will become \( \pi \) (which is not zero). \[ = \frac{\cos 0}{\pi-0} = \frac{1}{\pi} \]
In simple words: We just put 0 in place of \( x \). The top part becomes \( \cos 0 \) (which is 1), and the bottom part becomes \( \pi \). So the limit is \( \frac{1}{\pi} \).

Exam Tip: When evaluating limits of rational functions involving trigonometric terms, always check for direct substitution first. It's often the simplest approach if no indeterminate form arises.

Question 17. Evaluate the following limits: \( \lim_{x \rightarrow 0} \frac{\cos 2x-1}{\cos x-1} \)
Answer: Substituting \( x=0 \) directly yields \( \frac{\cos 0-1}{\cos 0-1} = \frac{1-1}{1-1} = \frac{0}{0} \), an indeterminate form. We use the trigonometric identity \( \cos 2x = 2\cos^2 x - 1 \). So, \( \cos 2x - 1 = (2\cos^2 x - 1) - 1 = 2\cos^2 x - 2 = 2(\cos^2 x - 1) \). This can be factored further using the difference of squares: \( 2(\cos x - 1)(\cos x + 1) \).
Now, substitute this back into the limit expression: \[ \lim_{x \rightarrow 0} \frac{2(\cos x - 1)(\cos x + 1)}{\cos x - 1} \] Since \( x \rightarrow 0 \), \( \cos x \ne 1 \), so \( (\cos x - 1) \) is not zero and we can cancel out the \( (\cos x - 1) \) terms: \[ = \lim_{x \rightarrow 0} 2(\cos x + 1) \] Now, substitute \( x=0 \): \[ = 2(\cos 0 + 1) = 2(1+1) = 2(2) = 4 \]
In simple words: This limit is \( \frac{0}{0} \) when you put in 0. We use a formula for \( \cos 2x \) to rewrite the top part. Then, we can cancel out the common \( (\cos x - 1) \) term from top and bottom. Finally, we put 0 back into the simpler expression to get 4.

Exam Tip: Look for opportunities to use double angle formulas or other trigonometric identities (like \( 1-\cos x = 2\sin^2 \frac{x}{2} \)) to simplify expressions involving trigonometric functions, especially when direct substitution results in an indeterminate form.

Question 18. Evaluate the following limits: \( \lim_{x \rightarrow 0} \frac{ax+x\cos x}{b\sin x} \)
Answer: Substituting \( x=0 \) yields \( \frac{0+0}{0} = \frac{0}{0} \), an indeterminate form. We can factor out \( x \) from the numerator: \[ \lim_{x \rightarrow 0} \frac{x(a+\cos x)}{b\sin x} \] Rearrange the terms to use the standard limit \( \lim_{\theta \rightarrow 0} \frac{\theta}{\sin \theta} = 1 \): \[ = \lim_{x \rightarrow 0} \left( \frac{1}{b} \cdot \frac{x}{\sin x} \cdot (a+\cos x) \right) \] Now, evaluate each part separately: \[ = \frac{1}{b} \cdot \left( \lim_{x \rightarrow 0} \frac{x}{\sin x} \right) \cdot \left( \lim_{x \rightarrow 0} (a+\cos x) \right) \] The first limit is 1, and for the third limit, we substitute \( x=0 \): \[ = \frac{1}{b} \cdot 1 \cdot (a+\cos 0) \] \[ = \frac{1}{b} \cdot (a+1) = \frac{a+1}{b} \]
In simple words: This limit is \( \frac{0}{0} \) when you put in 0. We take \( x \) out of the top part. Then, we rearrange the expression to use the standard rule \( \frac{x}{\sin x} \) becomes 1 as \( x \) goes to 0. Finally, we put 0 into the remaining parts to get the answer.

Exam Tip: When \( x \) approaches 0, look for \( \frac{x}{\sin x} \) or \( \frac{\sin x}{x} \) terms. Factoring out \( x \) from polynomial parts and strategically grouping terms can simplify the expression into these recognizable forms.

Question 19. Evaluate the following limits: \( \lim_{x \rightarrow 0} (x \sec x) \)
Answer: We can rewrite \( \sec x \) as \( \frac{1}{\cos x} \). So the limit becomes: \[ \lim_{x \rightarrow 0} \left( x \cdot \frac{1}{\cos x} \right) \] Now, we can substitute \( x=0 \) directly, as \( \cos 0 = 1 \), so the denominator will not be zero: \[ = \frac{0}{\cos 0} = \frac{0}{1} = 0 \]
In simple words: We change \( \sec x \) to \( \frac{1}{\cos x} \). Then, we can simply put 0 into the expression. Since \( \cos 0 \) is 1, the answer is 0.

Exam Tip: Always convert secant, cosecant, and cotangent functions into their sine and cosine equivalents when evaluating limits to simplify the process and identify potential indeterminate forms or direct substitutions.

Question 20. Evaluate the following limits: \( \lim_{x \rightarrow 0} \frac{\sin ax+bx}{ax+\sin bx}, a, b, a+b \ne 0 \)
Answer: Substituting \( x=0 \) directly yields \( \frac{0}{0} \), an indeterminate form. To handle this, we divide both the numerator and the denominator by \( x \): \[ \lim_{x \rightarrow 0} \frac{\frac{\sin ax}{x}+\frac{bx}{x}}{\frac{ax}{x}+\frac{\sin bx}{x}} \] \[ = \lim_{x \rightarrow 0} \frac{\frac{\sin ax}{x}+b}{a+\frac{\sin bx}{x}} \] Now, we can apply the limit to each term. Remember that \( \lim_{x \rightarrow 0} \frac{\sin kx}{kx} = 1 \implies \lim_{x \rightarrow 0} \frac{\sin kx}{x} = k \). So, \( \lim_{x \rightarrow 0} \frac{\sin ax}{x} = a \) and \( \lim_{x \rightarrow 0} \frac{\sin bx}{x} = b \). \[ = \frac{\lim_{x \rightarrow 0} (\frac{\sin ax}{x}) + \lim_{x \rightarrow 0} b}{\lim_{x \rightarrow 0} a + \lim_{x \rightarrow 0} (\frac{\sin bx}{x})} \] \[ = \frac{a+b}{a+b} \] Since \( a+b \ne 0 \), we can cancel the terms: \[ = 1 \]
In simple words: This limit is \( \frac{0}{0} \) if we put 0 in. We divide every part on the top and bottom by \( x \). Then we use the rule that \( \frac{\sin(\text{something})}{\text{something}} \) becomes 1 when the "something" goes to 0. After that, the expression simplifies to \( \frac{a+b}{a+b} \), which is 1.

Exam Tip: For limits involving sums of \( \sin x \) and \( x \) terms that result in \( \frac{0}{0} \), dividing all terms by \( x \) is a very effective strategy to apply the fundamental trigonometric limit properties.

Question 21. Evaluate the following limits: \( \lim_{x \rightarrow 0} (\csc x - \cot x) \)
Answer: Substituting \( x=0 \) directly leads to \( \csc 0 - \cot 0 \), which is \( \infty - \infty \), an indeterminate form. We rewrite \( \csc x \) and \( \cot x \) in terms of \( \sin x \) and \( \cos x \): \[ \lim_{x \rightarrow 0} \left( \frac{1}{\sin x} - \frac{\cos x}{\sin x} \right) \] Combine the fractions: \[ = \lim_{x \rightarrow 0} \frac{1-\cos x}{\sin x} \] This is still \( \frac{0}{0} \) if we substitute \( x=0 \). We can use the half-angle identities: \( 1-\cos x = 2\sin^2 (\frac{x}{2}) \) and \( \sin x = 2\sin (\frac{x}{2})\cos (\frac{x}{2}) \). \[ = \lim_{x \rightarrow 0} \frac{2\sin^2 (\frac{x}{2})}{2\sin (\frac{x}{2})\cos (\frac{x}{2})} \] Cancel out a common \( 2\sin (\frac{x}{2}) \) term: \[ = \lim_{x \rightarrow 0} \frac{\sin (\frac{x}{2})}{\cos (\frac{x}{2})} \] \[ = \lim_{x \rightarrow 0} \tan (\frac{x}{2}) \] Now, substitute \( x=0 \): \[ = \tan (\frac{0}{2}) = \tan 0 = 0 \]
In simple words: If we put 0 in, we get a tricky form. We change \( \csc x \) and \( \cot x \) into \( \sin x \) and \( \cos x \). Then we combine them into one fraction. To solve the remaining \( \frac{0}{0} \) form, we use special trigonometric formulas (half-angle identities) to simplify and then cancel terms. After that, we put 0 back in to get 0.

Exam Tip: When encountering \( \infty - \infty \) forms with trigonometric functions, convert all terms to sine and cosine. Then, use identities like \( 1-\cos x = 2\sin^2(x/2) \) or factor out common terms to simplify and resolve the indeterminate form.

Question 22. Evaluate the following limits: \( \lim_{x \rightarrow \frac{\pi}{2}} \frac{\tan 2x}{x-\frac{\pi}{2}} \)
Answer: If we substitute \( x=\frac{\pi}{2} \) directly, the numerator \( \tan(2 \cdot \frac{\pi}{2}) = \tan \pi = 0 \). The denominator \( \frac{\pi}{2}-\frac{\pi}{2} = 0 \). So it is \( \frac{0}{0} \), an indeterminate form. Let \( h = x-\frac{\pi}{2} \). Then \( x = \frac{\pi}{2}+h \). As \( x \rightarrow \frac{\pi}{2} \), \( h \rightarrow 0 \). Substitute \( h \) into the limit expression: \[ \lim_{h \rightarrow 0} \frac{\tan (2(\frac{\pi}{2}+h))}{h} \] \[ = \lim_{h \rightarrow 0} \frac{\tan (\pi+2h)}{h} \] Using the trigonometric identity \( \tan(\pi+\theta) = \tan \theta \): \[ = \lim_{h \rightarrow 0} \frac{\tan (2h)}{h} \] To use the standard limit \( \lim_{\theta \rightarrow 0} \frac{\tan \theta}{\theta} = 1 \), we multiply and divide by 2: \[ = \lim_{h \rightarrow 0} \frac{\tan (2h)}{2h} \cdot 2 \] \[ = \left( \lim_{h \rightarrow 0} \frac{\tan (2h)}{2h} \right) \cdot 2 \] Since \( 2h \rightarrow 0 \) as \( h \rightarrow 0 \), the first limit is 1. \[ = 1 \cdot 2 = 2 \]
In simple words: This limit is \( \frac{0}{0} \) if we put \( \frac{\pi}{2} \) in. We make a substitution by letting \( h = x-\frac{\pi}{2} \). This changes the limit to \( h \rightarrow 0 \). We use the trigonometric rule that \( \tan(\pi+\text{angle}) \) is the same as \( \tan(\text{angle}) \). Then, we use the special limit for \( \frac{\tan \theta}{\theta} \) to get the final answer.

Exam Tip: When the limit point is not 0 (e.g., \( x \rightarrow a \)), a common strategy is to make a substitution \( h = x-a \) or \( h = a-x \) so that the new variable \( h \) approaches 0. This allows the use of standard limits at 0.

Question 23. For the function \( f(x) = \left\{\begin{array}{l} 2x+3, x \leq 0 \\ 3(x+1), x>0 \end{array}\right. \), find \( \lim _{x \rightarrow 0^{-}} f(x) \), \( \lim _{x \rightarrow 0^{+}} f(x) \) and \( \lim _{x \rightarrow 0} f(x) \). Also, evaluate \( \lim _{x \rightarrow 1} f(x) \).
Answer:(i) To find the left-hand limit as \( x \rightarrow 0 \): Since \( x \leq 0 \), we use the function \( f(x) = 2x+3 \). \[ \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} (2x+3) = 2(0)+3 = 3 \] To find the right-hand limit as \( x \rightarrow 0 \): Since \( x > 0 \), we use the function \( f(x) = 3(x+1) \). \[ \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} 3(x+1) = 3(0+1) = 3(1) = 3 \] Since the left-hand limit equals the right-hand limit \( (3=3) \), the limit \( \lim_{x \rightarrow 0} f(x) \) exists and is equal to 3. \[ \lim_{x \rightarrow 0} f(x) = 3 \] (ii) To evaluate \( \lim_{x \rightarrow 1} f(x) \): Since \( x=1 \) is greater than 0, we use the function \( f(x) = 3(x+1) \) for \( x > 0 \). \[ \lim_{x \rightarrow 1} f(x) = \lim_{x \rightarrow 1} 3(x+1) = 3(1+1) = 3(2) = 6 \]
In simple words: For the limit at 0, we check the left side (where \( x \) is less than 0) and the right side (where \( x \) is greater than 0) of the function. Both sides give 3, so the limit at 0 is 3. For the limit at 1, since 1 is greater than 0, we use the second part of the function and put 1 in to get 6.

Exam Tip: For piecewise functions, always check the left-hand and right-hand limits at the boundary points to determine if the overall limit exists. If they are equal, the limit exists and equals that value.

Question 24. For the function \( f(x) = \left\{\begin{array}{l} x^{2}-1, x \leq 1 \\ -x^{2}-1, x>1 \end{array}\right. \), evaluate \( \lim_{x \rightarrow 1} f(x) \).
Answer: To find \( \lim_{x \rightarrow 1} f(x) \), we need to check the left-hand limit and the right-hand limit as \( x \) approaches 1. Left-hand limit: For \( x \leq 1 \), \( f(x) = x^2-1 \). \[ \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (x^2-1) = (1)^2-1 = 1-1 = 0 \] Right-hand limit: For \( x > 1 \), \( f(x) = -x^2-1 \). \[ \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1^{+}} (-x^2-1) = -(1)^2-1 = -1-1 = -2 \] Since the left-hand limit \( (0) \) is not equal to the right-hand limit \( (-2) \), the limit \( \lim_{x \rightarrow 1} f(x) \) does not exist.
In simple words: We look at the function's value as \( x \) gets close to 1 from the left and from the right. From the left, the value goes to 0. From the right, the value goes to -2. Because these two values are different, the limit at \( x=1 \) does not exist.

Exam Tip: If the left-hand limit and the right-hand limit of a piecewise function are not equal at a particular point, then the limit of the function at that point does not exist.

Question 25. Evaluate \( \lim _{x \rightarrow 0} f(x) \), where \( f(x) = \left\{\begin{array}{I} \frac{|x|}{x}, x \neq 0 \\ 0, \quad x=0 \end{array}\right. \).
Answer: To find \( \lim_{x \rightarrow 0} f(x) \), we evaluate the left-hand and right-hand limits. Left-hand limit: For \( x < 0 \), \( |x| = -x \). So, \( f(x) = \frac{-x}{x} = -1 \). \[ \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} (-1) = -1 \] Right-hand limit: For \( x > 0 \), \( |x| = x \). So, \( f(x) = \frac{x}{x} = 1 \). \[ \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} (1) = 1 \] Since the left-hand limit \( (-1) \) is not equal to the right-hand limit \( (1) \), the limit \( \lim_{x \rightarrow 0} f(x) \) does not exist.
In simple words: We check the limit as \( x \) approaches 0 from the left and from the right. When \( x \) is negative, \( \frac{|x|}{x} \) is -1. When \( x \) is positive, it's 1. Because these two values are different, the limit at \( x=0 \) doesn't exist.

Exam Tip: When evaluating limits involving the absolute value function, always split the problem into left-hand and right-hand limits to correctly evaluate \( |x| \) based on whether \( x \) is positive or negative.

Question 26. Find \( \lim_{x \rightarrow 0} f(x) \), where \( f(x) = \left\{\begin{array}{l} \frac{x}{|x|}, x \neq 0 \\ 0, \quad x=0 \end{array}\right. \).
Answer: To determine \( \lim_{x \rightarrow 0} f(x) \), we need to look at the left-hand limit and the right-hand limit. Left-hand limit: For \( x < 0 \), \( |x| = -x \). So, \( f(x) = \frac{x}{-x} = -1 \). \[ \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} (-1) = -1 \] Right-hand limit: For \( x > 0 \), \( |x| = x \). So, \( f(x) = \frac{x}{x} = 1 \). \[ \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} (1) = 1 \] Since the left-hand limit \( (-1) \) is not equal to the right-hand limit \( (1) \), the limit \( \lim_{x \rightarrow 0} f(x) \) does not exist.
In simple words: We check the limit as \( x \) approaches 0 from the left (negative values) and from the right (positive values). From the left, the function is -1. From the right, the function is 1. Since these are different, the limit at \( x=0 \) does not exist.

Exam Tip: The function \( \frac{x}{|x|} \) is a classic example of a function whose limit does not exist at \( x=0 \) due to different one-sided limits. Be ready to explain this concept clearly.

Question 27. Find \( \lim_{x \rightarrow 5} f(x) \), where \( f(x) = |x| – 5 \).
Answer: To find the limit, we evaluate the left-hand and right-hand limits at \( x=5 \). Left-hand limit: For \( x < 5 \), \( |x| = x \) (since \( x \) is positive). So, \( f(x) = x-5 \). \[ \lim_{x \rightarrow 5^{-}} f(x) = \lim_{x \rightarrow 5^{-}} (x-5) = 5-5 = 0 \] Right-hand limit: For \( x > 5 \), \( |x| = x \) (since \( x \) is positive). So, \( f(x) = x-5 \). \[ \lim_{x \rightarrow 5^{+}} f(x) = \lim_{x \rightarrow 5^{+}} (x-5) = 5-5 = 0 \] Since the left-hand limit equals the right-hand limit \( (0=0) \), the limit \( \lim_{x \rightarrow 5} f(x) \) exists and is equal to 0.
In simple words: We check the limit as \( x \) approaches 5 from both the left and the right. For values near 5, \( |x| \) is just \( x \). So, \( x-5 \) goes to 0 from both sides. This means the limit at \( x=5 \) is 0.

Exam Tip: For \( |x| \) functions, the absolute value sign can usually be removed for points not at 0 (or not where the argument of the absolute value is 0) by considering the sign of the argument in the interval. Here, for \( x \rightarrow 5 \), \( x \) is positive, so \( |x|=x \).

Question 28. Suppose \( f(x) = \left\{\begin{array}{l} a+b x, x<1 \\ 4, x=1 \\ b-a x, x>1 \end{array}\right. \) and if \( \lim_{x \rightarrow 1} f(x) = f(1) \), what are possible values of \( a \) and \( b \)?
Answer:We are given that \( f(1) = 4 \). For the limit \( \lim_{x \rightarrow 1} f(x) \) to exist and be equal to \( f(1) \), we must have: \( \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{+}} f(x) = f(1) \) First, calculate the left-hand limit: For \( x < 1 \), \( f(x) = a+bx \). \[ \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (a+bx) = a+b(1) = a+b \] Next, calculate the right-hand limit: For \( x > 1 \), \( f(x) = b-ax \). \[ \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1^{+}} (b-ax) = b-a(1) = b-a \] Now, set these limits equal to \( f(1) \): 1. \( a+b = 4 \) (from the left-hand limit) 2. \( b-a = 4 \) (from the right-hand limit) We now have a system of two linear equations: \( a+b = 4 \) ... (1) \( -a+b = 4 \) ... (2) Add equation (1) and equation (2): \( (a+b) + (-a+b) = 4+4 \) \( 2b = 8 \) \( b = 4 \) Substitute \( b=4 \) into equation (1): \( a+4 = 4 \) \( a = 0 \) So, the possible values are \( a=0 \) and \( b=4 \).
In simple words: We are told that the function's limit at 1 matches the function's value at 1, which is 4. This means the left-side limit and the right-side limit must both be 4. We set up equations for these limits using the given function parts. Solving these equations gives us \( a=0 \) and \( b=4 \).

Exam Tip: A function is continuous at a point if the limit exists at that point and equals the function's value. For piecewise functions, this means the left-hand limit, the right-hand limit, and the function value must all be equal at the point where the definition changes.

Question 29. Let \( a_1, a_2, \dots, a_n \) be fixed real numbers and define a function \( f(x) = (x – a_1)(x – a_2)\dots(x - a_n) \). What is \( \lim _{x \rightarrow a_{1}} f(x) \) and \( \lim _{x \rightarrow a} f(x) \)?
Answer:First, let's find \( \lim_{x \rightarrow a_1} f(x) \). We substitute \( x=a_1 \) into the function definition: \[ f(a_1) = (a_1 - a_1)(a_1 - a_2)\dots(a_1 - a_n) \] Since the first term \( (a_1 - a_1) \) is 0, the entire product becomes 0. \[ f(a_1) = 0 \cdot (a_1 - a_2)\dots(a_1 - a_n) = 0 \] Since \( f(x) \) is a polynomial function (it is a product of linear factors), it is continuous everywhere. Therefore, the limit as \( x \rightarrow a_1 \) is simply the function value at \( a_1 \). \[ \lim_{x \rightarrow a_1} f(x) = f(a_1) = 0 \] Next, let's find \( \lim_{x \rightarrow a} f(x) \) for any real number \( a \). Since \( f(x) \) is a polynomial, we can find the limit by direct substitution of \( x=a \) into the function. \[ \lim_{x \rightarrow a} f(x) = (a - a_1)(a - a_2)\dots(a - a_n) \] This expression is the value of the function \( f(a) \).
In simple words: For the limit as \( x \) approaches \( a_1 \), because \( (x-a_1) \) is a factor, putting \( x=a_1 \) into the function makes that factor zero, so the whole product becomes zero. For the limit as \( x \) approaches any number \( a \), we simply replace \( x \) with \( a \) in the function's formula.

Exam Tip: Polynomial functions are continuous everywhere. This means their limit at any point is simply the value of the function at that point. Factored forms clearly show the roots of the polynomial where the function's value (and thus the limit) is zero.

Question 30. If \( f(x) = \left\{\begin{array}{l} |x|+1, x<0 \\ 0, x=0 \\ |x|-1, x>0 \end{array}\right. \), discuss the limits at \( x=0 \) and at \( x=a \) for \( a \ne 0 \).
Answer:(i) Consider the limit at \( x=0 \): Left-hand limit: For \( x < 0 \), \( |x| = -x \). So \( f(x) = -x+1 \). \[ \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} (-x+1) = -(0)+1 = 1 \] Right-hand limit: For \( x > 0 \), \( |x| = x \). So \( f(x) = x-1 \). \[ \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} (x-1) = (0)-1 = -1 \] Since \( \lim_{x \rightarrow 0^{-}} f(x) = 1 \) and \( \lim_{x \rightarrow 0^{+}} f(x) = -1 \), the left-hand and right-hand limits are not equal. Therefore, \( \lim_{x \rightarrow 0} f(x) \) does not exist. (ii) Consider the limit at \( x=a \) where \( a \ne 0 \): Case 1: If \( a < 0 \). For values of \( x \) near \( a \) (where \( a < 0 \)), \( x \) will also be less than 0. So \( f(x) = |x|+1 = -x+1 \). \[ \lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} (-x+1) = -a+1 \] Case 2: If \( a > 0 \). For values of \( x \) near \( a \) (where \( a > 0 \)), \( x \) will also be greater than 0. So \( f(x) = |x|-1 = x-1 \). \[ \lim_{x \rightarrow a} f(x) = \lim_{x \rightarrow a} (x-1) = a-1 \] In summary: - At \( x=0 \), the limit does not exist. - At \( x=a \) for \( a<0 \), \( \lim_{x \rightarrow a} f(x) = -a+1 \). - At \( x=a \) for \( a>0 \), \( \lim_{x \rightarrow a} f(x) = a-1 \).
In simple words: For \( x=0 \), the limit from the left side is 1, but from the right side, it's -1. Since they are different, the limit at \( x=0 \) doesn't exist. For any other point \( a \), if \( a \) is negative, the limit is \( -a+1 \). If \( a \) is positive, the limit is \( a-1 \).

Exam Tip: When a function definition changes at a point (like \( x=0 \) here), it's crucial to check one-sided limits at that point. For other points where the function is defined by a single rule (e.g., \( x<0 \) or \( x>0 \)), the limit can usually be found by direct substitution within that rule.

Question 31. If the function \( f(x) \) satisfies \( \lim _{x \rightarrow 1} \frac{f(x)-2}{x^{2}-1} = \pi \), evaluate \( \lim _{x \rightarrow 1} f(x) \).
Answer: Let \( L = \lim_{x \rightarrow 1} f(x) \). We are given: \[ \lim_{x \rightarrow 1} \frac{f(x)-2}{x^2-1} = \pi \] We know that as \( x \rightarrow 1 \), the denominator \( x^2-1 \rightarrow 1^2-1 = 0 \). For the limit of a fraction to be a finite number \( \pi \) when the denominator approaches 0, the numerator must also approach 0. Therefore, we must have: \[ \lim_{x \rightarrow 1} (f(x)-2) = 0 \] Using the limit properties, this means: \[ \lim_{x \rightarrow 1} f(x) - \lim_{x \rightarrow 1} 2 = 0 \] \[ L - 2 = 0 \] \[ L = 2 \] So, \( \lim_{x \rightarrow 1} f(x) = 2 \).
In simple words: We are given that a fraction's limit is \( \pi \). Since the bottom part of the fraction goes to zero, the top part must also go to zero for the whole limit to be a real number. If \( f(x)-2 \) goes to 0, it means \( f(x) \) must go to 2.

Exam Tip: A key property of limits is that if \( \lim_{x \rightarrow c} \frac{g(x)}{h(x)} = L \) (where L is finite and non-zero) and \( \lim_{x \rightarrow c} h(x) = 0 \), then it must be true that \( \lim_{x \rightarrow c} g(x) = 0 \).

Question 32. For what integers \( m \) and \( n \) does both \( \lim _{x \rightarrow 0} f(x) \) and \( \lim _{x \rightarrow 1} f(x) \) exist for the following function? \( f(x) = \left\{\begin{array}{l} mx^2 + n, x < 0 \\ nx + m, 0 \leq x \leq 1 \\ nx^3 + m, x > 1 \end{array}\right. \)
Answer:For \( \lim_{x \rightarrow 0} f(x) \) to exist, the left-hand limit and the right-hand limit at \( x=0 \) must be equal. Left-hand limit at \( x=0 \): For \( x < 0 \), \( f(x) = mx^2+n \). \[ \lim_{x \rightarrow 0^{-}} f(x) = \lim_{x \rightarrow 0^{-}} (mx^2+n) = m(0)^2+n = n \] Right-hand limit at \( x=0 \): For \( 0 \leq x \leq 1 \), \( f(x) = nx+m \). \[ \lim_{x \rightarrow 0^{+}} f(x) = \lim_{x \rightarrow 0^{+}} (nx+m) = n(0)+m = m \] For the limit to exist at \( x=0 \), we must have \( n=m \).
For \( \lim_{x \rightarrow 1} f(x) \) to exist, the left-hand limit and the right-hand limit at \( x=1 \) must be equal. Left-hand limit at \( x=1 \): For \( 0 \leq x \leq 1 \), \( f(x) = nx+m \). \[ \lim_{x \rightarrow 1^{-}} f(x) = \lim_{x \rightarrow 1^{-}} (nx+m) = n(1)+m = n+m \] Right-hand limit at \( x=1 \): For \( x > 1 \), \( f(x) = nx^3+m \). \[ \lim_{x \rightarrow 1^{+}} f(x) = \lim_{x \rightarrow 1^{+}} (nx^3+m) = n(1)^3+m = n+m \] Since the left-hand limit \( (n+m) \) equals the right-hand limit \( (n+m) \), the limit \( \lim_{x \rightarrow 1} f(x) \) always exists for any integers \( m \) and \( n \). Therefore, the only condition for both limits to exist is \( m=n \). The integers \( m \) and \( n \) must be equal.
In simple words: For the limit at \( x=0 \) to exist, the function's value when \( x \) is just below 0 must match its value when \( x \) is just above 0. This means \( n \) must equal \( m \). For the limit at \( x=1 \), the value from the left and right sides are both \( n+m \), so this limit always exists. The only condition needed for both limits to exist is that \( m \) and \( n \) are the same number.

Exam Tip: When evaluating limits of piecewise functions at the points where their definitions change, compute both the left-hand and right-hand limits. For the limit to exist, these two limits must be equal, providing a condition on the parameters of the function.