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

Question 1. Lim \( x \rightarrow \infty \frac{4x-3}{2x+7} \)
Answer: We need to find the limit of the function as \( x \) approaches infinity. To do this, we can divide both the numerator and the denominator by the highest power of \( x \) in the denominator, which is \( x \).
\[ \text{Lt}_{x \rightarrow \infty} \frac{4x-3}{2x+7} = \text{Lt}_{x \rightarrow \infty} \frac{x(4-\frac{3}{x})}{x(2+\frac{7}{x})} \] Now, we can cancel out \( x \) from the numerator and denominator.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{4-\frac{3}{x}}{2+\frac{7}{x}} \] As \( x \) approaches infinity, \( \frac{3}{x} \) and \( \frac{7}{x} \) both approach 0.
\[ = \frac{4-0}{2+0} = \frac{4}{2} = 2 \] Alternatively, we can substitute \( x = \frac{1}{y} \). As \( x \rightarrow \infty \), \( y \rightarrow 0 \).
\[ \text{Lt}_{x \rightarrow \infty} \frac{4x-3}{2x+7} = \text{Lt}_{y \rightarrow 0} \frac{4(\frac{1}{y})-3}{2(\frac{1}{y})+7} = \text{Lt}_{y \rightarrow 0} \frac{\frac{4-3y}{y}}{\frac{2+7y}{y}} \] Cancel \( y \) from top and bottom.
\[ = \text{Lt}_{y \rightarrow 0} \frac{4-3y}{2+7y} \] Substitute \( y = 0 \).
\[ = \frac{4-3(0)}{2+7(0)} = \frac{4-0}{2+0} = \frac{4}{2} = 2 \]In simple words: To find the limit, we divide everything by 'x'. As 'x' gets very, very big, fractions like '3 divided by x' become zero. So, we are left with just the numbers.

🎯 Exam Tip: When evaluating limits at infinity for rational functions, divide the numerator and denominator by the highest power of x to simplify the expression. Remember that \( \frac{\text{constant}}{x^n} \) approaches 0 as \( x \rightarrow \infty \).

Question 2. Lim \( x \rightarrow \infty \frac{3x^{2}+2x-5}{x^{2}+5x+1} \)
Answer: To find the limit as \( x \) tends to infinity, divide every term in the numerator and denominator by the highest power of \( x \) present, which is \( x^2 \).
\[ \text{Lt}_{x \rightarrow \infty} \frac{3x^{2}+2x-5}{x^{2}+5x+1} = \text{Lt}_{x \rightarrow \infty} \frac{\frac{3x^{2}}{x^{2}}+\frac{2x}{x^{2}}-\frac{5}{x^{2}}}{\frac{x^{2}}{x^{2}}+\frac{5x}{x^{2}}+\frac{1}{x^{2}}} \] Simplify the fractions.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{3+\frac{2}{x}-\frac{5}{x^{2}}}{1+\frac{5}{x}+\frac{1}{x^{2}}} \] As \( x \) approaches infinity, terms like \( \frac{2}{x} \), \( \frac{5}{x^2} \), \( \frac{5}{x} \), and \( \frac{1}{x^2} \) all become 0. This is because a number divided by a very large number is almost zero.
\[ = \frac{3+0-0}{1+0+0} = \frac{3}{1} = 3 \]In simple words: When 'x' gets super big, only the terms with the highest power of 'x' really matter. Here, both top and bottom have 'x squared' as the biggest power. So, we just look at the numbers in front of those 'x squared' terms.

🎯 Exam Tip: For limits of rational functions where the degree of the numerator equals the degree of the denominator, the limit is the ratio of their leading coefficients. This is a quick way to check your answer.

Question 3. Lim \( x \rightarrow \infty \frac{x^{3}+6x^{2}+1}{x^{4}+3} \)
Answer: To evaluate this limit, we will divide every term in both the numerator and the denominator by the highest power of \( x \) found in the denominator, which is \( x^4 \).
\[ \text{Lt}_{x \rightarrow \infty} \frac{x^{3}+6x^{2}+1}{x^{4}+3} = \text{Lt}_{x \rightarrow \infty} \frac{\frac{x^{3}}{x^{4}}+\frac{6x^{2}}{x^{4}}+\frac{1}{x^{4}}}{\frac{x^{4}}{x^{4}}+\frac{3}{x^{4}}} \] Simplify the terms.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{\frac{1}{x}+\frac{6}{x^{2}}+\frac{1}{x^{4}}}{1+\frac{3}{x^{4}}} \] Now, as \( x \) approaches infinity, all terms with \( x \) in the denominator will approach 0.
\[ = \frac{0+0+0}{1+0} = \frac{0}{1} = 0 \]In simple words: In this problem, the bottom part (denominator) has a much higher power of 'x' than the top part (numerator). When this happens and 'x' gets very big, the whole fraction shrinks down to zero.

🎯 Exam Tip: If the degree of the denominator is greater than the degree of the numerator, the limit as \( x \rightarrow \infty \) will always be 0. This is a useful shortcut to remember.

Question 4. Lim \( x \rightarrow \infty \frac{3x^{3}+x^{2}-1}{x^{2}-x+7} \)
Answer: To find this limit, we divide all terms in the numerator and denominator by the highest power of \( x \) in the denominator, which is \( x^2 \).
\[ \text{Lt}_{x \rightarrow \infty} \frac{3x^{3}+x^{2}-1}{x^{2}-x+7} = \text{Lt}_{x \rightarrow \infty} \frac{\frac{3x^{3}}{x^{2}}+\frac{x^{2}}{x^{2}}-\frac{1}{x^{2}}}{\frac{x^{2}}{x^{2}}-\frac{x}{x^{2}}+\frac{7}{x^{2}}} \] Simplify the expression.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{3x+1-\frac{1}{x^{2}}}{1-\frac{1}{x}+\frac{7}{x^{2}}} \] As \( x \) approaches infinity, terms like \( \frac{1}{x^2} \) and \( \frac{1}{x} \) approach 0. However, the term \( 3x \) in the numerator approaches infinity.
\[ = \frac{3(\infty)+1-0}{1-0+0} = \frac{\infty}{1} = \infty \] Alternatively, let \( x = \frac{1}{y} \). As \( x \rightarrow \infty \), \( y \rightarrow 0 \).
\[ \text{Lt}_{x \rightarrow \infty} \frac{3x^{3}+x^{2}-1}{x^{2}-x+7} = \text{Lt}_{y \rightarrow 0} \frac{\frac{3}{y^{3}}+\frac{1}{y^{2}}-1}{\frac{1}{y^{2}}-\frac{1}{y}+7} \] Combine terms in the numerator and denominator by finding a common denominator.
\[ = \text{Lt}_{y \rightarrow 0} \frac{\frac{3+y-y^{3}}{y^{3}}}{\frac{1-y+7y^{2}}{y^{2}}} = \text{Lt}_{y \rightarrow 0} \frac{3+y-y^{3}}{y^{3}} \cdot \frac{y^{2}}{1-y+7y^{2}} \] Simplify the powers of \( y \).
\[ = \text{Lt}_{y \rightarrow 0} \frac{1}{y} \frac{3+y-y^{3}}{1-y+7y^{2}} \] Substitute \( y = 0 \).
\[ = \frac{1}{0} \frac{3+0-0}{1-0+0} = \frac{1}{0} \cdot 3 \] Since \( \frac{1}{0} \) is undefined and indicates an unbounded growth, the limit is infinity.
\[ = \infty \]In simple words: Here, the top part has a higher power of 'x' than the bottom part. When 'x' gets very, very large, the top grows much faster than the bottom, making the whole fraction grow without end, meaning the limit is infinity.

🎯 Exam Tip: If the degree of the numerator is greater than the degree of the denominator, the limit as \( x \rightarrow \infty \) will be either \( \infty \) or \( -\infty \), depending on the signs of the leading coefficients. Always check the leading coefficient signs.

Question 5. Lim \( x \rightarrow \infty \frac{5x-6}{\sqrt{4x^{2}+9}} \)
Answer: To find the limit as \( x \) approaches infinity, we divide every term in the numerator and denominator by the highest power of \( x \). In the denominator, \( \sqrt{4x^2} = 2x \), so the highest power is effectively \( x \).
\[ \text{Lt}_{x \rightarrow \infty} \frac{5x-6}{\sqrt{4x^{2}+9}} = \text{Lt}_{x \rightarrow \infty} \frac{x(5-\frac{6}{x})}{\sqrt{x^{2}(4+\frac{9}{x^{2}})}} \] Take \( x \) out of the square root in the denominator. Remember \( \sqrt{x^2} = x \) for positive \( x \).
\[ = \text{Lt}_{x \rightarrow \infty} \frac{x(5-\frac{6}{x})}{x\sqrt{4+\frac{9}{x^{2}}}} \] Cancel \( x \) from the numerator and denominator.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{5-\frac{6}{x}}{\sqrt{4+\frac{9}{x^{2}}}} \] As \( x \) approaches infinity, \( \frac{6}{x} \) and \( \frac{9}{x^2} \) both approach 0.
\[ = \frac{5-0}{\sqrt{4+0}} = \frac{5}{\sqrt{4}} = \frac{5}{2} \]In simple words: When you have a square root of 'x squared' in the bottom, it's like having 'x'. So, we divide everything by 'x'. As 'x' gets very big, all the small fractions become zero, leaving us with the main numbers.

🎯 Exam Tip: When dealing with square roots at infinity, divide by \( \sqrt{x^2} = x \) inside the root and \( x \) outside the root. Be careful with signs if \( x \rightarrow -\infty \).

Question 6. Lim \( x \rightarrow \infty \frac{\sqrt{3x^{2}-1}-\sqrt{2x^{2}-1}}{4x+3} \)
Answer: To find the limit as \( x \) approaches infinity, we divide every term by the highest power of \( x \). Here, the effective highest power is \( x \) (since \( \sqrt{x^2} = x \)).
First, take \( x^2 \) common from under each square root in the numerator.
\[ \text{Lt}_{x \rightarrow \infty} \frac{\sqrt{x^{2}(3-\frac{1}{x^{2}})}-\sqrt{x^{2}(2-\frac{1}{x^{2}})}}{x(4+\frac{3}{x})} \] Take \( x \) out of the square roots.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{x\sqrt{3-\frac{1}{x^{2}}}-x\sqrt{2-\frac{1}{x^{2}}}}{x(4+\frac{3}{x})} \] Factor out \( x \) from the numerator.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{x(\sqrt{3-\frac{1}{x^{2}}}-\sqrt{2-\frac{1}{x^{2}}})}{x(4+\frac{3}{x})} \] Cancel \( x \) from the numerator and denominator.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{\sqrt{3-\frac{1}{x^{2}}}-\sqrt{2-\frac{1}{x^{2}}}}{4+\frac{3}{x}} \] As \( x \) approaches infinity, \( \frac{1}{x^2} \) and \( \frac{3}{x} \) both approach 0.
\[ = \frac{\sqrt{3-0}-\sqrt{2-0}}{4+0} = \frac{\sqrt{3}-\sqrt{2}}{4} \]In simple words: We pull out 'x' from under the square roots on top and also from the bottom part. Then we cancel them out. As 'x' gets very big, the parts with 'x' at the bottom disappear, leaving just the square roots of the main numbers.

🎯 Exam Tip: When simplifying square roots like \( \sqrt{ax^2 \pm b} \), factor out \( x^2 \) to get \( x\sqrt{a \pm \frac{b}{x^2}} \). This helps identify the dominant terms for limits at infinity.

Question 7. Lim \( x \rightarrow \infty \sqrt{x} (\sqrt{x+c} - \sqrt{x}) \)
Answer: To evaluate this limit, we use the conjugate method to simplify the expression.
Multiply the term \( (\sqrt{x+c} - \sqrt{x}) \) by its conjugate \( (\sqrt{x+c} + \sqrt{x}) \) both in the numerator and denominator.
\[ \text{Lt}_{x \rightarrow \infty} \sqrt{x} (\sqrt{x+c} - \sqrt{x}) = \text{Lt}_{x \rightarrow \infty} \sqrt{x} (\sqrt{x+c} - \sqrt{x}) \cdot \frac{\sqrt{x+c} + \sqrt{x}}{\sqrt{x+c} + \sqrt{x}} \] Use the difference of squares formula, \( (a-b)(a+b) = a^2-b^2 \).
\[ = \text{Lt}_{x \rightarrow \infty} \frac{\sqrt{x}((x+c)-x)}{\sqrt{x+c} + \sqrt{x}} = \text{Lt}_{x \rightarrow \infty} \frac{c\sqrt{x}}{\sqrt{x+c} + \sqrt{x}} \] Now, divide the numerator and denominator by \( \sqrt{x} \).
\[ = \text{Lt}_{x \rightarrow \infty} \frac{c}{\frac{\sqrt{x+c}}{\sqrt{x}} + \frac{\sqrt{x}}{\sqrt{x}}} = \text{Lt}_{x \rightarrow \infty} \frac{c}{\sqrt{\frac{x+c}{x}} + 1} \] Simplify the term under the square root.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{c}{\sqrt{1+\frac{c}{x}} + 1} \] As \( x \) approaches infinity, \( \frac{c}{x} \) approaches 0.
\[ = \frac{c}{\sqrt{1+0} + 1} = \frac{c}{1+1} = \frac{c}{2} \]In simple words: We multiply the top and bottom by the "opposite sign" version of the square root part. This helps us get rid of the square roots in the numerator. After that, we divide by the biggest power of 'x' to find the final limit.

🎯 Exam Tip: When you have terms like \( \sqrt{A} - \sqrt{B} \), always think of multiplying by the conjugate \( \sqrt{A} + \sqrt{B} \) to rationalize the expression. This often simplifies the limit calculation significantly.

Question 8. Lim \( x \rightarrow \infty (\sqrt{x^{2}+x+1} - \sqrt{x^{2}+1}) \)
Answer: This limit requires using the conjugate to simplify the expression.
Multiply the expression by its conjugate \( (\sqrt{x^{2}+x+1} + \sqrt{x^{2}+1}) \) over itself.
\[ \text{Lt}_{x \rightarrow \infty} (\sqrt{x^{2}+x+1} - \sqrt{x^{2}+1}) = \text{Lt}_{x \rightarrow \infty} (\sqrt{x^{2}+x+1} - \sqrt{x^{2}+1}) \cdot \frac{\sqrt{x^{2}+x+1} + \sqrt{x^{2}+1}}{\sqrt{x^{2}+x+1} + \sqrt{x^{2}+1}} \] Apply the difference of squares formula, \( (a-b)(a+b) = a^2-b^2 \).
\[ = \text{Lt}_{x \rightarrow \infty} \frac{(x^{2}+x+1) - (x^{2}+1)}{\sqrt{x^{2}+x+1} + \sqrt{x^{2}+1}} \] Simplify the numerator.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{x}{\sqrt{x^{2}+x+1} + \sqrt{x^{2}+1}} \] Now, divide the numerator and denominator by the highest power of \( x \), which is \( x \). In the denominator, we take \( x^2 \) out of the square roots, so \( \sqrt{x^2} \) becomes \( x \).
\[ = \text{Lt}_{x \rightarrow \infty} \frac{\frac{x}{x}}{\sqrt{\frac{x^{2}}{x^{2}}+\frac{x}{x^{2}}+\frac{1}{x^{2}}} + \sqrt{\frac{x^{2}}{x^{2}}+\frac{1}{x^{2}}}} \] Simplify the terms.
\[ = \text{Lt}_{x \rightarrow \infty} \frac{1}{\sqrt{1+\frac{1}{x}+\frac{1}{x^{2}}} + \sqrt{1+\frac{1}{x^{2}}}} \] As \( x \) approaches infinity, terms like \( \frac{1}{x} \) and \( \frac{1}{x^2} \) approach 0.
\[ = \frac{1}{\sqrt{1+0+0} + \sqrt{1+0}} = \frac{1}{\sqrt{1} + \sqrt{1}} = \frac{1}{1+1} = \frac{1}{2} \]In simple words: This problem looks tricky because of the square roots. We use a trick called 'multiplying by the conjugate' to remove the square roots from the top. Then, we divide everything by 'x', and as 'x' gets very big, many parts become zero, leaving us with a simple fraction.

🎯 Exam Tip: For limits involving \( \sqrt{f(x)} - \sqrt{g(x)} \) at infinity, always use the conjugate. This method helps convert the expression into a rational function, which is easier to evaluate at infinity.

Question 9. Lim \( n \rightarrow \infty \frac{1+2+3+...+n}{n^{2}} \)
Answer: The sum of the first \( n \) natural numbers is given by the formula \( 1+2+3+...+n = \frac{n(n+1)}{2} \).
Substitute this sum into the limit expression.
\[ \text{given limit} = \text{Lt}_{n \rightarrow \infty} \frac{\sum n}{n^{2}} = \text{Lt}_{n \rightarrow \infty} \frac{\frac{n(n+1)}{2}}{n^{2}} \] Simplify the expression by rewriting the numerator.
\[ = \text{Lt}_{n \rightarrow \infty} \frac{n(n+1)}{2n^{2}} = \text{Lt}_{n \rightarrow \infty} \frac{n^{2}+n}{2n^{2}} \] Now, divide every term in the numerator and denominator by the highest power of \( n \), which is \( n^2 \).
\[ = \text{Lt}_{n \rightarrow \infty} \frac{\frac{n^{2}}{n^{2}}+\frac{n}{n^{2}}}{\frac{2n^{2}}{n^{2}}} = \text{Lt}_{n \rightarrow \infty} \frac{1+\frac{1}{n}}{2} \] As \( n \) approaches infinity, \( \frac{1}{n} \) approaches 0.
\[ = \frac{1+0}{2} = \frac{1}{2} \]In simple words: First, we use the formula to find the sum of numbers from 1 to 'n'. Then, we replace that sum in our problem. After that, we simplify and see what happens when 'n' gets very, very big, which makes the fraction \( \frac{1}{n} \) disappear.

🎯 Exam Tip: Remember the sum formulas for series, especially \( \sum n = \frac{n(n+1)}{2} \) and \( \sum n^2 = \frac{n(n+1)(2n+1)}{6} \), as they are frequently used in limits involving sums.

Question 10. Lim \( n \rightarrow \infty \frac{\sum n^{3}}{n^{4}} \)
Answer: The sum of the cubes of the first \( n \) natural numbers is given by the formula \( \sum n^{3} = \left(\frac{n(n+1)}{2}\right)^{2} = \frac{n^{2}(n+1)^{2}}{4} \).
Substitute this formula into the limit expression.
\[ \text{Lt}_{n \rightarrow \infty} \frac{\sum n^{3}}{n^{4}} = \text{Lt}_{n \rightarrow \infty} \frac{\frac{n^{2}(n+1)^{2}}{4}}{n^{4}} \] Simplify the expression.
\[ = \text{Lt}_{n \rightarrow \infty} \frac{n^{2}(n^{2}+2n+1)}{4n^{4}} = \text{Lt}_{n \rightarrow \infty} \frac{n^{4}+2n^{3}+n^{2}}{4n^{4}} \] Now, divide every term in the numerator and denominator by the highest power of \( n \), which is \( n^4 \).
\[ = \text{Lt}_{n \rightarrow \infty} \frac{\frac{n^{4}}{n^{4}}+\frac{2n^{3}}{n^{4}}+\frac{n^{2}}{n^{4}}}{\frac{4n^{4}}{n^{4}}} = \text{Lt}_{n \rightarrow \infty} \frac{1+\frac{2}{n}+\frac{1}{n^{2}}}{4} \] As \( n \) approaches infinity, terms like \( \frac{2}{n} \) and \( \frac{1}{n^2} \) approach 0. This is because a number divided by a very large number becomes very small.
\[ = \frac{1+0+0}{4} = \frac{1}{4} \]In simple words: First, we use the special formula for the sum of cubes of numbers up to 'n'. Then, we put this formula into our limit problem. We simplify the powers of 'n' and then see what happens as 'n' gets incredibly large.

🎯 Exam Tip: Being familiar with common series sum formulas, especially for \( \sum n \), \( \sum n^2 \), and \( \sum n^3 \), is crucial for solving limits involving sums. Always remember to divide by the highest power of the variable after substitution.