Class 11: Limits – Exercise 29.6 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Evaluate the following limits:

$\displaystyle \text{Question 1: } \lim \limits_{x \to \infty} \frac{(3x-1)(4x-2)}{(x+8)(x-1)}$

Answer:

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 2. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x^2.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0$

$\displaystyle \lim \limits_{x \to \infty} \frac{(3x-1)(4x-2)}{(x+8)(x-1)}$

$\displaystyle = \lim \limits_{x \to \infty} \frac{ \Big(\frac{3x-1}{x}\Big) \Big( \frac{4x-2}{x} \Big) }{ \Big( \frac{x+8}{x} \Big) \Big( \frac{x-1}{x} \Big)} = \lim \limits_{x \to \infty} \frac{ \Big(3 - \frac{1}{x}\Big) \Big( 4 - \frac{2}{x} \Big) }{ \Big( 1+ \frac{8}{x} \Big) \Big( 1 - \frac{1}{x} \Big)} = \frac{3 \times 4}{1} = 1$

$\\$

$\displaystyle \text{Question 2: } \lim \limits_{x \to \infty} \frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7}$

Answer:

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 3. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x^3.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x}, \frac{1}{x^2}, \frac{1}{x^3} \rightarrow 0$

$\displaystyle \lim \limits_{x \to \infty} \frac{3x^3-4x^2+6x-1}{2x^3+x^2-5x+7} = \lim \limits_{x \to \infty} \frac{3-\frac{4}{x}+\frac{6}{x^2}-1}{2+\frac{1}{x}-\frac{5}{x^2}+\frac{7}{x^3}} = \frac{3}{2}$

$\\$

$\displaystyle \text{Question 3: } \lim \limits_{x \to \infty} \frac{5x^3-6}{\sqrt{9+4x^6}}$

Answer:

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 3. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x^3.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x^3}, \frac{1}{x^6} \rightarrow 0$

$\displaystyle \lim \limits_{x \to \infty} \frac{5x^3-6}{\sqrt{9+4x^6}} = \lim \limits_{x \to \infty} \frac{\frac{5x^3-6}{x^3}}{\frac{\sqrt{9+4x^6}}{x^3}} = \lim \limits_{x \to \infty} \frac{5 - \frac{6}{x^3}}{\sqrt{\frac{9}{x^6}+4}} = \frac{5}{2}$

$\\$

$\displaystyle \text{Question 4: } \lim \limits_{x \to \infty} \sqrt{x^2+cx} -x$

Answer:

$\displaystyle \lim \limits_{x \to \infty} \sqrt{x^2+cx} -x$

This is of the form $\displaystyle \infty - \infty.$

Rationalizing the numerator:

$\displaystyle = \lim \limits_{x \to \infty} (\sqrt{x^2+cx} -x) \times \frac{\sqrt{x^2+cx} +x}{\sqrt{x^2+cx} +x}$

$\displaystyle = \lim \limits_{x \to \infty} \ \frac{x^2+cx - x^2}{\sqrt{x^2+cx} +x}$

$\displaystyle = \lim \limits_{x \to \infty} \ \frac{cx }{\sqrt{x^2+cx} +x}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0$

$\displaystyle = \lim \limits_{x \to \infty} \ \frac{c}{\sqrt{1+\frac{c}{x}} +1} = \frac{c}{\sqrt{1}+1} = \frac{c}{2}$

$\\$

$\displaystyle \text{Question 5: } \lim \limits_{x \to \infty} \sqrt{x+1}-\sqrt{x}$

Answer:

$\displaystyle \lim \limits_{x \to \infty} \sqrt{x+1}-\sqrt{x}$

This is of the form $\displaystyle \infty - \infty.$

Rationalizing the numerator:

$\displaystyle = \lim \limits_{x \to \infty} (\sqrt{x+1}-\sqrt{x}) \times \frac{\sqrt{x+1}+\sqrt{x}}{\sqrt{x+1}+\sqrt{x}}$

$\displaystyle = \lim \limits_{x \to \infty} \frac{x+1 - x }{\sqrt{x+1}+\sqrt{x}}$

$\displaystyle = \lim \limits_{x \to \infty} \frac{1 }{\sqrt{x+1}+\sqrt{x}}$

$\displaystyle = \frac{1}{\infty} = 0$

$\\$

$\displaystyle \text{Question 6: } \lim \limits_{x \to \infty} \sqrt{x^2+7x} - x$

Answer:

$\displaystyle \lim \limits_{x \to \infty} \sqrt{x^2+7x} - x$

This is of the form $\displaystyle \infty - \infty.$

Rationalizing the numerator:

$\displaystyle = \lim \limits_{x \to \infty} (\sqrt{x^2+7x} - x) \times \frac{\sqrt{x^2+7x} + x}{\sqrt{x^2+7x} + x}$

$\displaystyle = \lim \limits_{x \to \infty} \frac{x^2 + 7x - x^2}{\sqrt{x^2+7x} + x}$

$\displaystyle = \lim \limits_{x \to \infty} \frac{7x}{\sqrt{x^2+7x} + x}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0$

$\displaystyle = \lim \limits_{x \to \infty} \frac{7}{\sqrt{1+\frac{7}{x}} + 1}$

$\displaystyle = \frac{7}{\sqrt{1}+1} = \frac{7}{2}$

$\\$

$\displaystyle \text{Question 7: } \lim \limits_{x \to \infty} \frac{x}{\sqrt{4x^2+1}-1}$

Answer:

$\displaystyle \lim \limits_{x \to \infty} \frac{x}{\sqrt{4x^2+1}-1}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x}, \frac{1}{x^2} \rightarrow 0$

$\displaystyle = \lim \limits_{x \to \infty} \frac{1}{\sqrt{4+\frac{1}{x^2}}-\frac{1}{x}}$

$\displaystyle = \frac{1}{\sqrt{4}} = \frac{1}{2}$

$\\$

$\displaystyle \text{Question 8: } \lim \limits_{n \to \infty} \frac{n^2}{1+2+3+\cdots + n}$

Answer:

$\displaystyle \lim \limits_{n \to \infty} \frac{n^2}{1+2+3+\cdots + n} = \lim \limits_{n \to \infty} \frac{n^2}{\frac{n(n+1)}{2}}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ n \text{ in both the numerator and denominator is 2. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } n^2.$

$\displaystyle \text{When } n \rightarrow \infty, \text{ then } \frac{1}{n} \rightarrow 0$

$\displaystyle = \lim \limits_{n \to \infty} \frac{1}{\frac{1(1+\frac{1}{n})}{2}}$

$\displaystyle = \frac{2}{1} = 2$

$\\$

$\displaystyle \text{Question 9: } \lim \limits_{x \to \infty} \frac{3x^{-1}+4x^{-2}}{5x^{-1}+6x^{-2}}$

Answer:

$\displaystyle \lim \limits_{x \to \infty} \frac{3x^{-1}+4x^{-2}}{5x^{-1}+6x^{-2}} = \frac{\frac{3}{x}+ \frac{4}{x^2}}{\frac{5}{x}+\frac{6}{x^2}} = \frac{3x+4}{5x+6}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0$

$\displaystyle = \lim \limits_{x \to \infty} \frac{3+\frac{4}{x}}{5+\frac{6}{x}}$

$\displaystyle = \frac{3}{5}$

$\\$

$\displaystyle \text{Question 10: } \lim \limits_{x \to \infty} \frac{ \sqrt{x^2+a^2} - \sqrt{x^2+b^2} }{\sqrt{x^2+c^2} - \sqrt{x^2+d^2} }$

Answer:

$\displaystyle \lim \limits_{x \to \infty} \frac{ \sqrt{x^2+a^2} - \sqrt{x^2+b^2} }{\sqrt{x^2+c^2} - \sqrt{x^2+d^2} }$

Rationalizing the numerator and denominator we get:

$\displaystyle \lim \limits_{x \to \infty} \frac{ ( \sqrt{x^2+a^2} - \sqrt{x^2+b^2}) }{(\sqrt{x^2+c^2} - \sqrt{x^2+d^2} )} \times \frac{(\sqrt{x^2+c^2} + \sqrt{x^2+d^2} )}{(\sqrt{x^2+c^2} + \sqrt{x^2+d^2} )} \times \frac{( \sqrt{x^2+a^2} + \sqrt{x^2+b^2})}{( \sqrt{x^2+a^2} + \sqrt{x^2+b^2})}$

$\displaystyle \lim \limits_{x \to \infty} \frac{(x^2+a^2) - (x^2+b^2)}{(x^2+c^2) - (x^2+d^2)} \times \frac{(\sqrt{x^2+c^2} + \sqrt{x^2+d^2} )}{( \sqrt{x^2+a^2} + \sqrt{x^2+b^2})}$

$\displaystyle \lim \limits_{x \to \infty} \frac{a^2-b^2}{c^2-d^2} \times \frac{(\sqrt{x^2+c^2} + \sqrt{x^2+d^2} )}{( \sqrt{x^2+a^2} + \sqrt{x^2+b^2})}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0$

$\displaystyle \lim \limits_{x \to \infty} \frac{a^2-b^2}{c^2-d^2} \times \frac{(\sqrt{1+\frac{c^2}{x^2}} + \sqrt{1+\frac{d^2}{x^2}} )}{( \sqrt{1+\frac{a^2}{x^2}} + \sqrt{1+\frac{b^2}{x^2}})}$

$\displaystyle = \frac{a^2-b^2}{c^2-d^2} \times \frac{\sqrt{1}+\sqrt{1}}{\sqrt{1}+\sqrt{1}}$

$\displaystyle = \frac{a^2-b^2}{c^2-d^2}$

$\\$

$\displaystyle \text{Question 11: } \lim \limits_{n \to \infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}$

Answer:

$\displaystyle \lim \limits_{n \to \infty} \frac{(n+2)!+(n+1)!}{(n+2)!-(n+1)!}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{(n+2)(n+1)!+(n+1)!}{(n+2)(n+1)!-(n+1)!}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{n+3}{n+1} \times \frac{(n+1)!}{(n+1)!}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{n+3}{n+1}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ n \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } n.$

$\displaystyle \text{When } n \rightarrow \infty, \text{ then } \frac{1}{n} \rightarrow 0$

$\displaystyle = \lim \limits_{n \to \infty} \frac{1+\frac{3}{n}}{1+\frac{1}{n}}$

$\displaystyle = \frac{1}{1}$

$\displaystyle = 1$

$\\$

$\displaystyle \text{Question 12: } \lim \limits_{x \to \infty} x \Big\{ \sqrt{x^1+1}- \sqrt{x^2-1} \Big\}$

Answer:

$\displaystyle \lim \limits_{x \to \infty} x \Big\{ \sqrt{x^1+1}- \sqrt{x^2-1} \Big\}$

Rationalizing the numerator:

$\displaystyle = \lim \limits_{x \to \infty} x \Big\{ \sqrt{x^1+1}- \sqrt{x^2-1} \Big\} \times \frac{\sqrt{x^1+1}+ \sqrt{x^2-1}}{\sqrt{x^1+1}+ \sqrt{x^2-1}}$

$\displaystyle = \lim \limits_{x \to \infty} x \Bigg\{ \frac{(x^2+1)-(x^2-1)}{\sqrt{x^1+1}+ \sqrt{x^2-1}} \Bigg\}$

$\displaystyle = \lim \limits_{x \to \infty} \Bigg\{ \frac{2x}{\sqrt{x^1+1}+ \sqrt{x^2-1}} \Bigg\}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x}, \frac{1}{x^2} \rightarrow 0$

$\displaystyle = \lim \limits_{x \to \infty} \Bigg\{ \frac{2}{\sqrt{1+\frac{1}{x^2}}+ \sqrt{1-\frac{1}{x^2}}} \Bigg\}$

$\displaystyle = \frac{2}{\sqrt{1}+\sqrt{1}}$

$\displaystyle = 1$

$\\$

$\displaystyle \text{Question 13: } \lim \limits_{x \to \infty} \Big\{ \sqrt{x+1} - \sqrt{x} \Big\} \sqrt{x+2}$

Answer:

$\displaystyle \lim \limits_{x \to \infty} \Big\{ \sqrt{x+1} - \sqrt{x} \Big\} \sqrt{x+2}$

Rationalizing the numerator:

$\displaystyle = \lim \limits_{x \to \infty} \Big\{ \sqrt{x+1} - \sqrt{x} \Big\} \sqrt{x+2} \times \frac{\sqrt{x+1} + \sqrt{x}}{\sqrt{x+1} + \sqrt{x}}$

$\displaystyle = \lim \limits_{x \to \infty} \Bigg\{ \frac{(\sqrt{x+2}(x+1-x)}{\sqrt{x+1} + \sqrt{x}} \Bigg\}$

$\displaystyle = \lim \limits_{x \to \infty} \Bigg\{ \frac{\sqrt{x+2}}{\sqrt{x+1} + \sqrt{x}} \Bigg\}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 1/2. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } \sqrt{x}.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0$

$\displaystyle = \lim \limits_{x \to \infty} \Bigg\{ \frac{\sqrt{1+\frac{2}{x}}}{\sqrt{1+\frac{1}{x}} + \sqrt{1}} \Bigg\}$

$\displaystyle = \frac{1}{\sqrt{1}+1} = \frac{1}{2}$

$\\$

$\displaystyle \text{Question 14: } \lim \limits_{n \to \infty} \frac{1^2+2^2+ \cdots+ n^2}{n^3}$

Answer:

$\displaystyle \lim \limits_{n \to \infty} \frac{1^2+2^2+ \cdots+ n^2}{n^3}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{n(n+1)(2n+1)}{6n^3}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}$

$\displaystyle = \frac{2}{6} = \frac{1}{3}$

$\\$

$\displaystyle \text{Question 15: } \lim \limits_{n \to \infty} \Bigg( \frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + \cdots + \frac{n-1}{n^2} \Bigg)$

Answer:

$\displaystyle \lim \limits_{n \to \infty} \Bigg( \frac{1}{n^2} + \frac{2}{n^2} + \frac{3}{n^2} + \cdots + \frac{n-1}{n^2} \Bigg)$

$\displaystyle = \lim \limits_{n \to \infty} \Bigg( \frac{1+2+3+ \cdots + (n-1)}{n^2} \Bigg)$

$\displaystyle = \lim \limits_{n \to \infty} \Bigg( \frac{n(n-1)}{2n^2} \Bigg)$

$\displaystyle = \lim \limits_{n \to \infty} \Bigg( \frac{1(1-\frac{1}{n})}{2} \Bigg)$

$\displaystyle = \frac{1}{2}$

$\\$

$\displaystyle \text{Question 16: } \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots+ n^3}{n^4}$

Answer:

$\displaystyle \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots+ n^3}{n^4}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{ \Big( \frac{n(n+1)}{2} \Big)^2 }{n^4}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{ \frac{n^2(n+1)^2}{4} }{n^4}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{n^2}{n^2} \times \frac{(n+1)^2}{4n^2}$

$\displaystyle = \lim \limits_{n \to \infty} \Big(1 + \frac{1}{n} \Big)^2 \times \frac{1}{4}$

$\displaystyle = \frac{1}{4}$

$\\$

$\displaystyle \text{Question 17: } \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots+ n^3}{(n-1)^4}$

Answer:

$\displaystyle \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots+ n^3}{(n-1)^4}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{ \Big( \frac{n(n+1)}{2} \Big)^2 }{(n-1)^4}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{ n^2(n+1)^2 }{ 4(n-1)^4 }$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ n \text{ in both the numerator and denominator is 4. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } n^4.$

$\displaystyle \text{When } n \rightarrow \infty, \text{ then } \frac{1}{n} \rightarrow 0$

$\displaystyle = \lim \limits_{n \to \infty} \frac{ 1(1+\frac{1}{n})^2 }{ 4(1-\frac{1}{n})^4 }$

$\displaystyle = \frac{1}{4}$

$\\$

$\displaystyle \text{Question 18: } \lim \limits_{x \to \infty} \sqrt{x} \Big\{ \sqrt{x+1}-\sqrt{x} \Big\}$

Answer:

$\displaystyle \lim \limits_{x \to \infty} \sqrt{x} \Big\{ \sqrt{x+1}-\sqrt{x} \Big\}$

$\displaystyle = \lim \limits_{x \to \infty} \sqrt{x} \Bigg\{ (\sqrt{x+1}-\sqrt{x} ) \times \frac{ \sqrt{x+1}+\sqrt{x}}{ \sqrt{x+1}+\sqrt{x}} \Bigg\}$

$\displaystyle = \lim \limits_{x \to \infty} \Bigg\{ \frac{\sqrt{x} (x+1-x)}{\sqrt{x+1}+\sqrt{x}} \Bigg \}$

$\displaystyle = \lim \limits_{x \to \infty} \Bigg\{ \frac{\sqrt{x} }{\sqrt{x+1}+\sqrt{x}} \Bigg \}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 1/2. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } \sqrt{x}.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} \rightarrow 0$

$\displaystyle = \lim \limits_{x \to \infty} \Bigg\{ \frac{1 }{\sqrt{1+\frac{1}{x}}+1} \Bigg \}$

$\displaystyle = \frac{1}{2}$

$\\$

$\displaystyle \text{Question 19: } \lim \limits_{n \to \infty} \Bigg( \frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots + \frac{1}{3^n} \Bigg)$

Answer:

$\displaystyle \lim \limits_{n \to \infty} \Bigg( \frac{1}{3^1} + \frac{1}{3^2} + \frac{1}{3^3} + \cdots + \frac{1}{3^n} \Bigg)$

$\displaystyle = \lim \limits_{n \to \infty} \Bigg[ \frac{1}{3} \Bigg( 1 + \frac{1}{3^1} + \frac{1}{3^2} + \cdots + \frac{1}{3^(n-1) } \Bigg) \Bigg]$

$\displaystyle = \lim \limits_{n \to \infty} \Bigg[ \frac{\frac{1}{3} \Big( 1-\frac{1}{3^n} \Big)}{1 - \frac{1}{3}} \Bigg]$

$\displaystyle = \lim \limits_{n \to \infty} \frac{1}{2} \Bigg( 1 - \frac{1}{3^n} \Bigg)$

$\displaystyle = \frac{1}{2} (1-0) = \frac{1}{2}$

$\\$

$\displaystyle \text{Question 20: } \lim \limits_{x \to \infty} \frac{x^4+7x^3+46x+a}{x^4+6} , \text{ where a is a non-zero real number}$

Answer:

$\displaystyle \lim \limits_{x \to \infty} \frac{x^4+7x^3+46x+a}{x^4+6}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 4. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x^4.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} , \frac{1}{x^2}, \frac{1}{x^3} , \frac{1}{x^4} \rightarrow 0$

$\displaystyle = \lim \limits_{x \to \infty} \frac{1+\frac{7}{x}+\frac{46}{x^3}+\frac{a}{x^4} }{1+\frac{6}{x^4} }$

$\displaystyle = \frac{1}{1} = 1$

$\\$

$\displaystyle \text{Question 21: } f(x) = \frac{ax^2+b}{x^2+1}, \lim \limits_{x \to 0} f(x) = 1 \text{ and } \lim \limits_{x \to \infty} f(x) = 1, \\ \\ \text{ then prove that } f(-2) = f(2) = 1.$

Answer:

$\displaystyle f(x) = \frac{ax^2+b}{x^2+1}$

$\displaystyle \lim \limits_{x \to 0} f(x) = 1$

$\displaystyle \Rightarrow \lim \limits_{x \to 0} \frac{ax^2+b}{x^2+1} = 1$

$\displaystyle \Rightarrow \frac{a \times 0 + b}{0+1} =1$

$\displaystyle \Rightarrow b = 1$

$\displaystyle \text{Also } \lim \limits_{x \to \infty} f(x) = 1$

$\displaystyle \Rightarrow \lim \limits_{x \to \infty}\frac{ax^2+b}{x^2+1} = 1$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} , \frac{1}{x^2} \rightarrow 0$

$\displaystyle \Rightarrow \lim \limits_{x \to \infty}\frac{a+\frac{b}{x^2}}{1+\frac{1}{x^2}} = 1$

$\displaystyle \Rightarrow \frac{a+0}{1+0} = 1$

$\displaystyle \Rightarrow a = 1$

$\displaystyle \Rightarrow f(x) = \frac{x^2+1}{x^2+1} = 1$

$\displaystyle f(x) \text{ is a constant function and does not depend on } x.$

$\displaystyle \text{Therefore } f(-2) = 1 \text{ and } f(2) = 1$

$\\$

$\displaystyle \text{Question 22: } \text{Show that } \lim \limits_{x \to \infty} ( \sqrt{x^2+x+1} - 1) \neq \lim \limits_{x \to \infty} ( \sqrt{x^2+1} - x)$

Answer:

LHS:

$\displaystyle \lim \limits_{x \to \infty} ( \sqrt{x^2+x+1} - x)$

$\displaystyle = \lim \limits_{x \to \infty} ( \sqrt{x^2+x+1} - x) \times \frac{\sqrt{x^2+x+1} + x}{\sqrt{x^2+x+1} + x}$

$\displaystyle = \lim \limits_{x \to \infty} \frac{x^2+x+1-x^2}{\sqrt{x^2+x+1} + x}$

$\displaystyle = \lim \limits_{x \to \infty} \frac{x+1}{\sqrt{x^2+x+1} + x}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ x \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } x.$

$\displaystyle \text{When } x \rightarrow \infty, \text{ then } \frac{1}{x} , \frac{1}{x^2} \rightarrow 0$

$\displaystyle = \lim \limits_{x \to \infty} \frac{1+\frac{1}{x}}{\sqrt{1+\frac{1}{x}+\frac{1}{x^2}} + 1}$

$\displaystyle = \frac{1}{\sqrt{1}+1} = \frac{1}{2}$

RHS:

$\displaystyle \lim \limits_{x \to \infty} ( \sqrt{x^2+1} - x)$

$\displaystyle = \lim \limits_{x \to \infty} ( \sqrt{x^2+1} - x) \times \frac{\sqrt{x^2+1} +x}{\sqrt{x^2+1} +x}$

$\displaystyle = \lim \limits_{x \to \infty} \frac{x^2+1-x^2}{\sqrt{x^2+1} +x}$

$\displaystyle = \lim \limits_{x \to \infty} \frac{1}{\sqrt{x^2+1} +x}$

$\displaystyle = \frac{1}{\infty} = 0$

$\displaystyle \text{Therefore } \lim \limits_{x \to \infty} ( \sqrt{x^2+x+1} - 1) \neq \lim \limits_{x \to \infty} ( \sqrt{x^2+1} - x)$

$\\$

$\displaystyle \text{Question 23: } \lim \limits_{x \to - \infty} ( \sqrt{ 4x^2-7x} +2x)$

Answer:

$\displaystyle \text{Given: } \lim \limits_{x \to - \infty} ( \sqrt{ 4x^2-7x} +2x)$

$\displaystyle \text{Let } x = - m. \text{ When } x \rightarrow - \infty, \text{ then } m \rightarrow \infty .$

$\displaystyle \Rightarrow \lim \limits_{m \to \infty} ( \sqrt{ 4m^2+7m} -2m)$

$\displaystyle = \lim \limits_{m \to \infty} ( \sqrt{ 4m^2+7m} -2m) \times \frac{\sqrt{ 4m^2+7m} + 2m}{\sqrt{ 4m^2+7m} +2m}$

$\displaystyle = \lim \limits_{m \to \infty} \frac{4m^2+7m-4m^2}{\sqrt{ 4m^2+7m} + 2m}$

$\displaystyle = \lim \limits_{m \to \infty} \frac{7m}{\sqrt{ 4m^2+7m} + 2m}$

$\displaystyle \text{Here the expression assumes the form } \frac{\infty}{\infty}. \text{ We notice that the highest power of} \\ \\ m \text{ in both the numerator and denominator is 1. Therefore, we divide each term } \\ \\ \text{ in both the numerator and denominator by } m.$

$\displaystyle \text{When } m \rightarrow \infty, \text{ then } \frac{1}{m} \rightarrow 0$

$\displaystyle = \lim \limits_{m \to \infty} \frac{7}{\sqrt{ 4+\frac{7}{m}} + 2}$

$\displaystyle = \frac{7}{\sqrt{4}+2} = \frac{7}{4}$

$\\$

$\displaystyle \text{Question 24: } \lim \limits_{x \to - \infty} ( \sqrt{ x^2-8x} +x)$

Answer:

$\displaystyle \text{Given: } \lim \limits_{x \to - \infty} ( \sqrt{ x^2-8x} +x)$

$\displaystyle \text{Let } x = - m. \text{ When } x \rightarrow - \infty, \text{ then } m \rightarrow \infty .$

$\displaystyle = \lim \limits_{m \to \infty} ( \sqrt{ m^2+8m} -m )$

$\displaystyle = \lim \limits_{m \to \infty} ( \sqrt{ m^2+8m} -m ) \times \frac{\sqrt{ m^2+8m} +m }{\sqrt{ m^2+8m} + m }$

$\displaystyle = \lim \limits_{m \to \infty} \frac{m^2+8m - m^2}{\sqrt{ m^2+8m} + m}$

$\displaystyle = \lim \limits_{m \to \infty} \frac{8m}{\sqrt{ m^2+8m} + m}$

$\displaystyle \text{When } m \rightarrow \infty, \text{ then } \frac{1}{m} \rightarrow 0$

$\displaystyle = \lim \limits_{m \to \infty} \frac{8}{\sqrt{ 1+\frac{8}{m}} + 1}$

$\displaystyle = \frac{8}{\sqrt{1}+1} = 4$

$\\$

$\displaystyle \text{Question 25: } \text{Evaluate: } \lim \limits_{n \to \infty} \frac{1^4+2^4+3^4+ \cdots + n^4}{n^5} - \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots + n^3}{n^5}$

Answer:

$\displaystyle \text{Consider the identity } (k+1)^5-k^5 = 5k^4+10k^3+10k^2+5k+1$

$\displaystyle \text{Putting } k = 1, 2, 3 , \cdots , n \text{ and then adding the equations we have }$

$\displaystyle (n+1)^4 - 1 = 5 \sum_{k=1}^{n} k^4 + 10 \sum_{k=1}^{n} k^3 + 10 \sum_{k=1}^{n} k^2 + 5 \sum_{k=1}^{n} k + \sum_{k=1}^{n}$

$\Rightarrow n^5 + 5n^4 + 10n^3 + 10n^2 + 5n = 5 \sum_{k=1}^{n}+ \frac{10n^2(n+1)^2}{4} + \frac{10n(n+1)(2n+1)}{6}+ \frac{5n(n+1)}{2} + n$

$\Rightarrow 5 \sum_{k=1}^{n} k^4 = n^5+5n^4+10n^3+10n^3+4n - \frac{5n^2(n+1)^2}{2} - \frac{5n(n+1)(2n+1)}{3}- \frac{5n(n+1)}{2}$

$\displaystyle \Rightarrow 5 \sum_{k=1}^{n} k^4 = n^5 + \frac{5n^4}{2} + \frac{5n^3}{3}- \frac{n}{6}$

This expression on further simplification gives

$\displaystyle \sum_{k=1}^{n} k^4 = \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$

$\displaystyle \therefore \lim \limits_{n \to \infty} \frac{1^4+2^4+3^4+ \cdots + n^4}{n^5} - \lim \limits_{n \to \infty} \frac{1^3+2^3+ \cdots + n^3}{n^5}$

$\displaystyle = \lim \limits_{n \to \infty} \frac{n(n+1)(2n+1)(3n^2+3n-1)}{30n^5} - \lim \limits_{n \to \infty} \frac{n^2(n+1)^2}{4n^5}$

$\displaystyle = \frac{1}{30} \lim \limits_{n \to \infty} \Bigg( 1 + \frac{1}{n} \Bigg) \Bigg( 2 + \frac{1}{n} \Bigg) \Bigg(3+\frac{3}{n}- \frac{1}{n^2} \Bigg) - \frac{1}{4} \lim \limits_{n \to \infty} \frac{1}{n} \Bigg(1 + \frac{1}{n} \Bigg)^2$