OP Malhotra Class 11 Maths Solutions Chapter 18 Limits Exercise 18 (C)

S Chand Class 11 ICSE Maths Solutions Chapter 18 Limits Ex 18(c)

Question 1. \( \text{Lim}_{x \to -1} \frac{x^2-1}{x+1} \)
Answer: To solve this limit problem, we first factorize the numerator. The expression \( x^2-1 \) can be written as \( (x-1)(x+1) \). Then, we cancel out the common term \( (x+1) \) from both the numerator and the denominator. After that, we substitute the value \( x = -1 \) into the simplified expression \( (x-1) \). This gives us \( -1-1 \), which equals \( -2 \). Factoring helps eliminate the indeterminate form before substitution.
\( \text{Lt}_{x \to -1} \frac{x^2-1}{x+1} = \text{Lt}_{x \to -1} \frac{(x-1)(x+1)}{x+1} \)
\( = \text{Lt}_{x \to -1} (x-1) = -1-1 = -2 \)
In simple words: First, we break down \( x^2-1 \) into \( (x-1)(x+1) \). Then, we cross out \( (x+1) \) from the top and bottom. Finally, we put \( x = -1 \) into what's left, \( (x-1) \), which gives us \( -2 \).

🎯 Exam Tip: Always look for ways to factorize and cancel common terms before substituting the limit value to avoid indeterminate forms.

Question 2. \( \text{Lim}_{x \to 1} \frac{(2x-3)(x-1)}{2x^2+x-3} \)
Answer: For this limit problem, we start by factoring the denominator. The expression \( 2x^2+x-3 \) can be factored into \( (x-1)(2x+3) \). We can then cancel out the common factor \( (x-1) \) from both the numerator and the denominator. After simplifying, we substitute the value \( x = 1 \) into the remaining expression \( \frac{2x-3}{2x+3} \). This calculation leads to \( \frac{2(1)-3}{2(1)+3} \), which simplifies to \( \frac{-1}{5} \). Factoring is crucial here to avoid division by zero.
\( \text{Lt}_{x \to 1} \frac{(2x-3)(x-1)}{2x^2+x-3} = \text{Lt}_{x \to 1} \frac{(2x-3)(x-1)}{(x-1)(2x+3)} \)
\( = \text{Lt}_{x \to 1} \frac{2x-3}{2x+3} = \frac{2 \times 1 - 3}{2 \times 1 + 3} = \frac{-1}{5} \)
In simple words: First, we factor the bottom part of the fraction to get \( (x-1)(2x+3) \). Then, we cross out \( (x-1) \) from top and bottom. Next, we put \( x = 1 \) into the new fraction. The final answer is \( \frac{-1}{5} \).

🎯 Exam Tip: When dealing with rational functions, always factorize the denominator to identify and cancel any common factors that might cause the expression to be undefined at the limit point.

Question 3. \( \text{Lim}_{x \to -1} \frac{(x-4)(x+1)}{x^2+3x+2} \)
Answer: To find this limit, we first factor the denominator \( x^2+3x+2 \) into \( (x+1)(x+2) \). Then, we cancel the common factor \( (x+1) \) that appears in both the numerator and the denominator. After this simplification, we substitute \( x = -1 \) into the expression \( \frac{x-4}{x+2} \). Performing the calculation, we get \( \frac{-1-4}{-1+2} \), which simplifies to \( \frac{-5}{1} \), resulting in \( -5 \). This method helps resolve the indeterminate form.
\( \text{Lt}_{x \to -1} \frac{(x-4)(x+1)}{x^2+3x+2} = \text{Lt}_{x \to -1} \frac{(x-4)(x+1)}{(x+1)(x+2)} \)
\( = \text{Lt}_{x \to -1} \frac{x-4}{x+2} = \frac{-1-4}{-1+2} = \frac{-5}{1} = -5 \)
In simple words: First, factor the bottom part into \( (x+1)(x+2) \). Then, cancel \( (x+1) \) from the top and bottom. Put \( x = -1 \) into the new fraction. The answer is \( -5 \).

🎯 Exam Tip: Always double-check your factorization steps, especially for quadratic expressions, as a small error can lead to a completely different result.

Question 4. \( \text{Lim}_{x \to \frac{1}{2}} \frac{4x^2-1}{2x-1} \)
Answer: To calculate this limit, we recognize that the numerator \( 4x^2-1 \) is a difference of squares, which can be factored as \( (2x-1)(2x+1) \). We then cancel the common term \( (2x-1) \) from both the numerator and the denominator. Finally, we substitute \( x = \frac{1}{2} \) into the simplified expression \( (2x+1) \). This yields \( 2 \left( \frac{1}{2} \right) + 1 \), which simplifies to \( 1+1=2 \). This factorization is key to solving the limit.
\( \text{Lt}_{x \to \frac{1}{2}} \frac{4x^2-1}{2x-1} = \text{Lt}_{x \to \frac{1}{2}} \frac{(2x-1)(2x+1)}{2x-1} \)
\( = \text{Lt}_{x \to \frac{1}{2}} (2x+1) = 2 \left( \frac{1}{2} \right) + 1 = 1+1 = 2 \)
In simple words: We can write \( 4x^2-1 \) as \( (2x-1)(2x+1) \). Then, we can cross out \( (2x-1) \) from the top and bottom. After that, we put \( x = \frac{1}{2} \) into \( (2x+1) \), and the answer is \( 2 \).

🎯 Exam Tip: Remember the difference of squares formula, \( a^2-b^2 = (a-b)(a+b) \), as it is very useful for simplifying limit expressions.

Question 5. \( \text{Lim}_{x \to 2} \frac{7x^2-11x-6}{3x^2-x-10} \)
Answer: First, we factorize both the numerator and the denominator. The numerator \( 7x^2-11x-6 \) factors into \( (x-2)(7x+3) \), and the denominator \( 3x^2-x-10 \) factors into \( (x-2)(3x+5) \). Then, we cancel the common factor \( (x-2) \) from both the top and bottom. After simplifying, we substitute \( x=2 \) into the expression \( \frac{7x+3}{3x+5} \). This gives us \( \frac{7(2)+3}{3(2)+5} \), which calculates to \( \frac{14+3}{6+5} = \frac{17}{11} \). This technique effectively resolves the indeterminate form.
\( \text{Lt}_{x \to 2} \frac{7x^2-11x-6}{3x^2-x-10} \) (This is in `\( \frac{0}{0} \)` form.)
\( = \text{Lt}_{x \to 2} \frac{(x-2)(7x+3)}{(x-2)(3x+5)} \)
\( = \text{Lt}_{x \to 2} \frac{7x+3}{3x+5} \)
\( = \frac{7 \times 2 + 3}{3 \times 2 + 5} = \frac{14+3}{6+5} = \frac{17}{11} \)
In simple words: We factor the top part into \( (x-2)(7x+3) \) and the bottom part into \( (x-2)(3x+5) \). Then, we cancel out \( (x-2) \). Finally, we put \( x=2 \) into the new fraction to get \( \frac{17}{11} \).

🎯 Exam Tip: For quadratic factorizations like \( ax^2+bx+c \), if you know one root (e.g., \( x=2 \) in this case), then \( (x-2) \) must be a factor, which simplifies finding the other factor.

Question 6. \( \text{Lim}_{x \to 2} \frac{x^2(x^2-4)}{x-2} \)
Answer: To evaluate this limit, we first factor the term \( (x^2-4) \) in the numerator as a difference of squares, which becomes \( (x-2)(x+2) \). Then, we cancel the common factor \( (x-2) \) from both the numerator and the denominator. After this simplification, we substitute \( x=2 \) into the remaining expression \( x^2(x+2) \). This calculation gives us \( 2^2(2+2) \), which is \( 4 \times 4 \), resulting in \( 16 \). Factoring helps remove the part that causes an undefined value.
\( \text{Lt}_{x \to 2} \frac{x^2(x^2-4)}{x-2} = \text{Lt}_{x \to 2} \frac{x^2(x-2)(x+2)}{x-2} \)
\( = \text{Lt}_{x \to 2} x^2(x+2) = 2^2(2+2) = 4 \times 4 = 16 \)
In simple words: First, we break \( (x^2-4) \) into \( (x-2)(x+2) \). Then, we cross out \( (x-2) \) from top and bottom. Next, we put \( x=2 \) into what's left, \( x^2(x+2) \), to get \( 16 \).

🎯 Exam Tip: Always fully factorize all polynomial terms, especially if they are a difference of squares or cubes, to simplify the expression before substituting the limit value.

Question 7. \( \text{Lim}_{x \to 2} \left( \frac{x^8-16}{x^4-4} + \frac{x^2-9}{x-3} \right) \)
Answer: To solve this problem, we evaluate each part of the sum separately. For the first fraction, we factor the numerator \( (x^8-16) \) as \( (x^4-4)(x^4+4) \), and then cancel the common factor \( (x^4-4) \) with the denominator. For the second fraction, we factor \( (x^2-9) \) as \( (x-3)(x+3) \) and cancel \( (x-3) \). After these simplifications, we are left with \( (x^4+4) + (x+3) \). Substituting \( x=2 \) into this expression gives us \( (2^4+4) + (2+3) \), which calculates to \( (16+4) + 5 \), resulting in \( 20+5=25 \). Handling each term separately makes complex limits easier.
\( \text{Lt}_{x \to 2} \left( \frac{x^8-16}{x^4-4} + \frac{x^2-9}{x-3} \right) \)
\( = \text{Lt}_{x \to 2} \left( \frac{(x^4-4)(x^4+4)}{x^4-4} + \frac{(x-3)(x+3)}{x-3} \right) \)
\( = \text{Lt}_{x \to 2} \left( x^4+4 + x+3 \right) \)
\( = (2^4+4) + (2+3) \)
\( = (16+4) + (5) \)
\( = 20 + 5 = 25 \)
In simple words: We break down the first fraction: \( x^8-16 \) becomes \( (x^4-4)(x^4+4) \). Cancel \( (x^4-4) \). For the second fraction, \( x^2-9 \) becomes \( (x-3)(x+3) \). Cancel \( (x-3) \). What's left is \( (x^4+4) + (x+3) \). Now, put \( x=2 \) into this to get \( (16+4) + (2+3) = 20+5 = 25 \).

🎯 Exam Tip: When evaluating limits of sums or differences, apply the limit operator to each term separately, provided each individual limit exists.

Question 8. \( \text{Lim}_{x \to 2} \frac{x^3-8}{x-2} \)
Answer: To solve this limit, we use the algebraic identity for the difference of cubes, where \( x^3-8 \) can be written as \( x^3-2^3 \). This factors into \( (x-2)(x^2+2x+4) \). We then cancel the common factor \( (x-2) \) from the numerator and denominator. Finally, we substitute \( x=2 \) into the simplified expression \( (x^2+2x+4) \). This calculation gives us \( 2^2+2(2)+4 \), which simplifies to \( 4+4+4=12 \). Knowing common factoring formulas helps simplify such expressions.
\( \text{Lt}_{x \to 2} \frac{x^3-8}{x-2} \)
\( = \text{Lt}_{x \to 2} \frac{(x-2)(x^2+2x+4)}{x-2} \)
\( = \text{Lt}_{x \to 2} (x^2+2x+4) \)
\( = 2^2 + 2(2) + 4 = 4+4+4 = 12 \)
In simple words: We use a special formula for \( x^3-8 \) which turns it into \( (x-2)(x^2+2x+4) \). Then, we cross out \( (x-2) \) from the top and bottom. Next, we put \( x=2 \) into \( (x^2+2x+4) \), and the answer is \( 12 \).

🎯 Exam Tip: Memorize the difference of cubes formula: \( a^3-b^3 = (a-b)(a^2+ab+b^2) \). It's a common tool for simplifying limits involving cubic expressions.

Question 9. \( \text{Lim}_{x \to \frac{1}{2}} \left( \frac{8x-3}{2x-1} - \frac{4x^2+1}{4x^2-1} \right) \)
Answer: First, we find a common denominator for the two fractions, which is \( (2x-1)(2x+1) \), as \( 4x^2-1 \) is \( (2x-1)(2x+1) \). We then combine the fractions. After simplifying the numerator, we get \( 12x^2+2x-4 \). We can factor this numerator as \( 2(6x^2+x-2) \), and further as \( 2(2x-1)(3x+2) \). We then cancel the common factor \( (2x-1) \) from the numerator and denominator. Finally, we substitute \( x = \frac{1}{2} \) into the simplified expression \( \frac{2(3x+2)}{2x+1} \). This leads to \( \frac{2(3 \times \frac{1}{2} + 2)}{2 \times \frac{1}{2} + 1} \), which simplifies to \( \frac{7}{2} \). Common denominators and factorization are essential for these types of limits.
\( \text{Lt}_{x \to \frac{1}{2}} \left( \frac{8x-3}{2x-1} - \frac{4x^2+1}{4x^2-1} \right) \)
\( = \text{Lt}_{x \to \frac{1}{2}} \left( \frac{8x-3}{2x-1} - \frac{4x^2+1}{(2x-1)(2x+1)} \right) \)
\( = \text{Lt}_{x \to \frac{1}{2}} \frac{(8x-3)(2x+1) - (4x^2+1)}{(2x-1)(2x+1)} \)
\( = \text{Lt}_{x \to \frac{1}{2}} \frac{(16x^2+8x-6x-3) - (4x^2+1)}{(2x-1)(2x+1)} \)
\( = \text{Lt}_{x \to \frac{1}{2}} \frac{16x^2+2x-3-4x^2-1}{(2x-1)(2x+1)} \)
\( = \text{Lt}_{x \to \frac{1}{2}} \frac{12x^2+2x-4}{(2x-1)(2x+1)} \)
\( = \text{Lt}_{x \to \frac{1}{2}} \frac{2(6x^2+x-2)}{(2x-1)(2x+1)} \)
\( = \text{Lt}_{x \to \frac{1}{2}} \frac{2(2x-1)(3x+2)}{(2x-1)(2x+1)} \)
\( = \text{Lt}_{x \to \frac{1}{2}} \frac{2(3x+2)}{2x+1} \)
\( = \frac{2(3 \times \frac{1}{2} + 2)}{2 \times \frac{1}{2} + 1} = \frac{2(\frac{3}{2}+2)}{1+1} = \frac{2(\frac{3+4}{2})}{2} = \frac{2(\frac{7}{2})}{2} = \frac{7}{2} \)
In simple words: First, we make the bottom parts of the fractions the same. We combine the fractions and simplify the top part. Then, we factor the top part again to find common factors. We cross out the common \( (2x-1) \). Finally, we put \( x = \frac{1}{2} \) into the final simple fraction to get \( \frac{7}{2} \).

🎯 Exam Tip: When faced with an indeterminate form of type \( \infty - \infty \), combine the fractions using a common denominator to transform it into a \( \frac{0}{0} \) form, which can then be solved by factorization.

Question 10. \( \text{Lim}_{x \to 3} \frac{x^3-8x^2+45}{2x^2-3x-9} \)
Answer: To find this limit, we first factorize both the numerator \( x^3-8x^2+45 \) into \( (x-3)(x^2-5x-15) \) and the denominator \( 2x^2-3x-9 \) into \( (x-3)(2x+3) \). We then cancel the common factor \( (x-3) \) from both parts of the fraction. After simplifying, we substitute \( x=3 \) into the expression \( \frac{x^2-5x-15}{2x+3} \). This leads to \( \frac{3^2 - 5(3) - 15}{2(3)+3} \), which simplifies to \( \frac{9-15-15}{6+3} = \frac{-21}{9} \). This fraction can be further reduced to \( -\frac{7}{3} \). Factorization helps eliminate the zero in the denominator.
\( \text{Lt}_{x \to 3} \frac{x^3-8x^2+45}{2x^2-3x-9} \) (This is in `\( \frac{0}{0} \)` form.)
\( = \text{Lt}_{x \to 3} \frac{(x-3)(x^2-5x-15)}{(x-3)(2x+3)} \)
\( = \text{Lt}_{x \to 3} \frac{x^2-5x-15}{2x+3} \)
\( = \frac{3^2 - 5 \times 3 - 15}{2 \times 3 + 3} \)
\( = \frac{9-15-15}{6+3} \)
\( = \frac{-21}{9} = -\frac{7}{3} \)
In simple words: We factor the top part as \( (x-3)(x^2-5x-15) \) and the bottom part as \( (x-3)(2x+3) \). We then cross out \( (x-3) \). Finally, we put \( x=3 \) into the remaining fraction. The answer is \( -\frac{7}{3} \).

🎯 Exam Tip: For higher-degree polynomials that result in \( \frac{0}{0} \) form, if \( x=a \) makes the expression zero, then \( (x-a) \) is a factor. You can use synthetic division or polynomial long division to find the other factors.

Question 11. \( \text{Lim}_{x \to 3} \frac{x^3-6x-9}{x^4-81} \)
Answer: First, we factorize the numerator \( x^3-6x-9 \) as \( (x-3)(x^2+3x+3) \). For the denominator, \( x^4-81 \), we factor it as a difference of squares \( (x^2-9)(x^2+9) \), and then further factor \( (x^2-9) \) into \( (x-3)(x+3) \). So, the denominator becomes \( (x-3)(x+3)(x^2+9) \). We then cancel the common factor \( (x-3) \) from the top and bottom. After simplifying, we substitute \( x=3 \) into the expression \( \frac{x^2+3x+3}{(x+3)(x^2+9)} \). This calculation gives us \( \frac{3^2+3(3)+3}{(3+3)(3^2+9)} \), which simplifies to \( \frac{9+9+3}{(6)(9+9)} = \frac{21}{6 \times 18} = \frac{21}{108} \). Finally, we reduce the fraction to its simplest form, \( \frac{7}{36} \). Knowing how to factor various polynomial forms is crucial here.
\( \text{Lt}_{x \to 3} \frac{x^3-6x-9}{x^4-81} \) (This is in `\( \frac{0}{0} \)` form.)
\( = \text{Lt}_{x \to 3} \frac{(x-3)(x^2+3x+3)}{(x^2-9)(x^2+9)} \)
\( = \text{Lt}_{x \to 3} \frac{(x-3)(x^2+3x+3)}{(x-3)(x+3)(x^2+9)} \)
\( = \text{Lt}_{x \to 3} \frac{x^2+3x+3}{(x+3)(x^2+9)} \)
\( = \frac{3^2+3(3)+3}{(3+3)(3^2+9)} \)
\( = \frac{9+9+3}{(6)(9+9)} = \frac{21}{6 \times 18} = \frac{21}{108} \)
\( = \frac{7}{36} \)
In simple words: We factor the top part and the bottom part. The bottom part \( x^4-81 \) needs to be factored twice. Then, we cross out the common factor \( (x-3) \). After that, we put \( x=3 \) into the new fraction to get \( \frac{21}{108} \), which simplifies to \( \frac{7}{36} \).

🎯 Exam Tip: When dealing with higher powers, remember the difference of squares formula can be applied multiple times (e.g., \( x^4-81 = (x^2-9)(x^2+9) \) and then \( x^2-9 = (x-3)(x+3) \)).

Question 12. \( \text{Lim}_{x \to \sqrt{2}} \frac{x^4-4}{x^2+3x\sqrt{2}-8} \)
Answer: To solve this limit, we factor both the numerator and the denominator. The numerator \( x^4-4 \) can be written as \( x^4-(\sqrt{2})^4 \), which factors into \( (x^2-2)(x^2+2) \), and further into \( (x-\sqrt{2})(x+\sqrt{2})(x^2+2) \). The denominator \( x^2+3x\sqrt{2}-8 \) factors into \( (x-\sqrt{2})(x+4\sqrt{2}) \). We then cancel the common factor \( (x-\sqrt{2}) \) from the top and bottom. After simplifying, we substitute \( x=\sqrt{2} \) into the remaining expression \( \frac{(x+\sqrt{2})(x^2+2)}{x+4\sqrt{2}} \). This gives us \( \frac{(\sqrt{2}+\sqrt{2})((\sqrt{2})^2+2)}{\sqrt{2}+4\sqrt{2}} \), which simplifies to \( \frac{(2\sqrt{2})(2+2)}{5\sqrt{2}} = \frac{8\sqrt{2}}{5\sqrt{2}} \). Finally, canceling \( \sqrt{2} \) gives us \( \frac{8}{5} \). Factoring polynomials with radical roots requires careful attention.
\( \text{Lt}_{x \to \sqrt{2}} \frac{x^4-4}{x^2+3x\sqrt{2}-8} \)
\( = \text{Lt}_{x \to \sqrt{2}} \frac{x^4-(\sqrt{2})^4}{x^2+3x\sqrt{2}-8} \)
\( = \text{Lt}_{x \to \sqrt{2}} \frac{(x^2-(\sqrt{2})^2)(x^2+(\sqrt{2})^2)}{(x-\sqrt{2})(x+4\sqrt{2})} \)
\( = \text{Lt}_{x \to \sqrt{2}} \frac{(x-\sqrt{2})(x+\sqrt{2})(x^2+2)}{(x-\sqrt{2})(x+4\sqrt{2})} \)
\( = \text{Lt}_{x \to \sqrt{2}} \frac{(x+\sqrt{2})(x^2+2)}{x+4\sqrt{2}} \)
\( = \frac{(\sqrt{2}+\sqrt{2})((\sqrt{2})^2+2)}{\sqrt{2}+4\sqrt{2}} \)
\( = \frac{(2\sqrt{2})(2+2)}{5\sqrt{2}} = \frac{2\sqrt{2} \times 4}{5\sqrt{2}} \)
\( = \frac{8\sqrt{2}}{5\sqrt{2}} = \frac{8}{5} \)
In simple words: We factor the top part \( x^4-4 \) as \( (x-\sqrt{2})(x+\sqrt{2})(x^2+2) \). We factor the bottom part as \( (x-\sqrt{2})(x+4\sqrt{2}) \). Then, we cross out \( (x-\sqrt{2}) \). Now, we put \( x=\sqrt{2} \) into the new fraction to get \( \frac{(2\sqrt{2})(4)}{5\sqrt{2}} \), which simplifies to \( \frac{8}{5} \).

🎯 Exam Tip: When evaluating limits with irrational numbers, remember that \( x^2-a = (x-\sqrt{a})(x+\sqrt{a}) \) can be a useful factorization for both numerator and denominator.

Question 13. \( \text{Lim}_{x \to 1} \frac{1-x^{-\frac{1}{3}}}{1-x^{-\frac{2}{3}}} \)
Answer: To evaluate this limit, we first rewrite the denominator \( 1-x^{-\frac{2}{3}} \) as a difference of squares: \( 1-(x^{-\frac{1}{3}})^2 \). This allows us to factor it into \( (1-x^{-\frac{1}{3}})(1+x^{-\frac{1}{3}}) \). Then, we cancel the common factor \( (1-x^{-\frac{1}{3}}) \) from both the numerator and the denominator. After simplifying, we substitute \( x=1 \) into the remaining expression \( \frac{1}{1+x^{-\frac{1}{3}}} \). Since any power of 1 is 1, this simplifies to \( \frac{1}{1+1} = \frac{1}{2} \). Recognizing patterns like difference of squares is helpful here.
\( \text{Lt}_{x \to 1} \frac{1-x^{-\frac{1}{3}}}{1-x^{-\frac{2}{3}}} \)
\( = \text{Lt}_{x \to 1} \frac{1-x^{-\frac{1}{3}}}{1-(x^{-\frac{1}{3}})^2} \)
\( = \text{Lt}_{x \to 1} \frac{1-x^{-\frac{1}{3}}}{(1-x^{-\frac{1}{3}})(1+x^{-\frac{1}{3}})} \)
\( = \text{Lt}_{x \to 1} \frac{1}{1+x^{-\frac{1}{3}}} \)
\( = \frac{1}{1+1^{-\frac{1}{3}}} = \frac{1}{1+1} = \frac{1}{2} \)
In simple words: First, we change the bottom part \( 1-x^{-\frac{2}{3}} \) to \( 1-(x^{-\frac{1}{3}})^2 \). Then, we factor it as \( (1-x^{-\frac{1}{3}})(1+x^{-\frac{1}{3}}) \). We cross out \( (1-x^{-\frac{1}{3}}) \). Finally, we put \( x=1 \) into the simple fraction \( \frac{1}{1+x^{-\frac{1}{3}}} \), which gives us \( \frac{1}{1+1} = \frac{1}{2} \).

🎯 Exam Tip: Look for opportunities to apply algebraic identities to terms with fractional exponents, as they often simplify complex expressions. Always remember that \( 1^n = 1 \) for any real number \( n \).

Question 14. \( \text{Lim}_{x \to 2} \frac{x-2}{\sqrt{x}-\sqrt{2}} \)
Answer: To solve this limit, we notice that the numerator \( x-2 \) can be written as a difference of squares involving square roots: \( (\sqrt{x})^2 - (\sqrt{2})^2 \). This factors into \( (\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2}) \). We then cancel the common factor \( (\sqrt{x}-\sqrt{2}) \) from both the numerator and the denominator. After this simplification, we substitute \( x=2 \) into the remaining expression \( (\sqrt{x}+\sqrt{2}) \). This results in \( \sqrt{2}+\sqrt{2} \), which equals \( 2\sqrt{2} \). Transforming integer terms into squares of their roots is a useful trick for these limits.
\( \text{Lt}_{x \to 2} \frac{x-2}{\sqrt{x}-\sqrt{2}} \)
\( = \text{Lt}_{x \to 2} \frac{(\sqrt{x})^2 - (\sqrt{2})^2}{\sqrt{x}-\sqrt{2}} \)
\( = \text{Lt}_{x \to 2} \frac{(\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2})}{\sqrt{x}-\sqrt{2}} \)
\( = \text{Lt}_{x \to 2} (\sqrt{x}+\sqrt{2}) \)
\( = \sqrt{2}+\sqrt{2} = 2\sqrt{2} \)
In simple words: We can write the top part \( x-2 \) as \( (\sqrt{x})^2 - (\sqrt{2})^2 \). This factors into \( (\sqrt{x}-\sqrt{2})(\sqrt{x}+\sqrt{2}) \). We then cross out \( (\sqrt{x}-\sqrt{2}) \). Finally, we put \( x=2 \) into \( (\sqrt{x}+\sqrt{2}) \), which gives \( 2\sqrt{2} \).

🎯 Exam Tip: When you see expressions like \( x-a \) and \( \sqrt{x}-\sqrt{a} \), remember that \( x-a \) can be factored as \( (\sqrt{x}-\sqrt{a})(\sqrt{x}+\sqrt{a}) \) to simplify the limit.

Question 15. If \( \text{Lim}_{x \to 1} \frac{x^4-1}{x-1} = \text{Lim}_{x \to k} \frac{x^3-k^3}{x^2-k^2}, \text{ find the value of } k. \)
Answer: We are given an equation where two limits are equal. First, let's solve the Left Hand Side (LHS) limit. We factor the numerator \( x^4-1 \) as \( (x^2-1)(x^2+1) \), and then further as \( (x-1)(x+1)(x^2+1) \). After canceling \( (x-1) \) from the numerator and denominator, we substitute \( x=1 \), which gives us \( (1+1)(1^2+1) = 2 \times 2 = 4 \). Next, we solve the Right Hand Side (RHS) limit. We factor the numerator \( x^3-k^3 \) using the difference of cubes formula and the denominator \( x^2-k^2 \) using the difference of squares formula. This results in \( \frac{(x-k)(x^2+kx+k^2)}{(x-k)(x+k)} \). After canceling \( (x-k) \), we substitute \( x=k \), which gives us \( \frac{k^2+k(k)+k^2}{k+k} = \frac{3k^2}{2k} = \frac{3k}{2} \). Now, we set the LHS and RHS equal: \( 4 = \frac{3k}{2} \). Multiplying both sides by 2 gives \( 8 = 3k \), so \( k = \frac{8}{3} \). We also check if \( k=0 \) is a possible solution, but it is not, as it would make the RHS equal to 0, which does not match the LHS value of 4. This problem combines factoring techniques with solving an algebraic equation.
Given \( \text{Lt}_{x \to 1} \frac{x^4-1}{x-1} = \text{Lt}_{x \to k} \frac{x^3-k^3}{x^2-k^2} \)
For LHS:
\( \text{Lt}_{x \to 1} \frac{x^4-1}{x-1} = \text{Lt}_{x \to 1} \frac{(x^2-1)(x^2+1)}{x-1} \)
\( = \text{Lt}_{x \to 1} \frac{(x-1)(x+1)(x^2+1)}{x-1} \)
\( = \text{Lt}_{x \to 1} (x+1)(x^2+1) \)
\( = (1+1)(1^2+1) = 2 \times 2 = 4 \)
For RHS:
\( \text{Lt}_{x \to k} \frac{x^3-k^3}{x^2-k^2} = \text{Lt}_{x \to k} \frac{(x-k)(x^2+kx+k^2)}{(x-k)(x+k)} \)
\( = \text{Lt}_{x \to k} \frac{x^2+kx+k^2}{x+k} \)
\( = \frac{k^2+k(k)+k^2}{k+k} = \frac{3k^2}{2k} = \frac{3k}{2} \) (assuming \( k \neq 0 \))
Equating LHS and RHS:
\( 4 = \frac{3k}{2} \)
\( \implies 8 = 3k \)
\( \implies k = \frac{8}{3} \)
If \( k=0 \), then RHS \( = \text{Lt}_{x \to 0} \frac{x^3}{x^2} = \text{Lt}_{x \to 0} x = 0 \), which is not equal to LHS \( = 4 \). So, \( k=0 \) is not a solution.
Thus, the required value of \( k \) is \( \frac{8}{3} \).
In simple words: We need to find a value for 'k' that makes both sides of the equation true. For the left side, we factor \( x^4-1 \) and simplify, then put \( x=1 \), which gives \( 4 \). For the right side, we factor \( x^3-k^3 \) and \( x^2-k^2 \), simplify, and then put \( x=k \), which gives \( \frac{3k}{2} \). Setting \( 4 = \frac{3k}{2} \), we find \( k = \frac{8}{3} \). We also make sure that \( k=0 \) doesn't work, because it would give \( 0 \) on the right side, not \( 4 \).

🎯 Exam Tip: When a problem involves limits of equal expressions, evaluate each limit separately first, then equate the results to solve for the unknown variable. Remember to consider all possible values and edge cases for the variable.