Class 11: Limits – Exercise 29.5 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Note:

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We have,

$\displaystyle \lim \limits_{x \to a} \frac{x^m-a^m}{x^n-a^n}$

$\displaystyle = \lim \limits_{x \to a} \Bigg\{ \frac{x^m-a^m}{x-a} \times \frac{x-a}{x^n-a^n} \Bigg\}$

$\displaystyle = \lim \limits_{x \to a} \Bigg\{ \frac{x^m-a^m}{x-a} \div \frac{x^n-a^n}{x-a} \Bigg\}$

$\displaystyle = \lim \limits_{x \to a} \frac{x^m-a^m}{x-a} \div \lim \limits_{x \to a} \frac{x^n-a^n}{x-a}$

$\displaystyle = ma^{m-1} \div na^{n-1}$

$\displaystyle = \frac{m}{n} a^{m-n}$

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$\displaystyle \text{Question 1: } \lim \limits_{x \to a} \frac{(x+2)^\frac{5}{2} - ( a+2)^\frac{5}{2}}{x-a }$

Answer:

We have,

$\displaystyle \lim \limits_{x \to a} \frac{(x+2)^\frac{5}{2} - ( a+2)^\frac{5}{2}}{x-a }$

$\displaystyle \text{Let } y = x+2 \text{ and } b = a+2$

$\displaystyle \text{When } x \rightarrow a, \text{ then } x+2 \rightarrow a+2 \Rightarrow y \rightarrow b$

$\displaystyle = \lim \limits_{y \to b} \frac{y^\frac{5}{2} - b^\frac{5}{2}}{y-b }$

$\displaystyle = \frac{5}{2} b^{\frac{5}{2}-1}$

$\displaystyle = \frac{5}{2} b^{\frac{3}{2}}$

$\displaystyle = \frac{5}{2} (a+2)^{\frac{3}{2}}$

$\\$

$\displaystyle \text{Question 2: } \lim \limits_{x \to a} \frac{(x+2)^\frac{3}{2} - ( a+2)^\frac{3}{2}}{x-a }$

Answer:

We have,

$\displaystyle \lim \limits_{x \to a} \frac{(x+2)^\frac{3}{2} - ( a+2)^\frac{3}{2}}{x-a }$

$\displaystyle \text{Let } y = x+2 \text{ and } b = a+2$

$\displaystyle \text{When } x \rightarrow a, \text{ then } x+2 \rightarrow a+2 \Rightarrow y \rightarrow b$

$\displaystyle = \lim \limits_{y \to b} \frac{y^\frac{3}{2} - b^\frac{3}{2}}{y-b }$

$\displaystyle = \frac{3}{2} b^{\frac{3}{2}-1}$

$\displaystyle = \frac{3}{2} b^{\frac{1}{2}}$

$\displaystyle = \frac{3}{2} (a+2)^{\frac{1}{2}}$

$\\$

$\displaystyle \text{Question 3: } \lim \limits_{x \to 0} \frac{(1+x)^6-1}{(1+x)^2-1 }$

Answer:

We have,

$\displaystyle \lim \limits_{x \to 0} \frac{(1+x)^6-1}{(1+x)^2-1 }$

$\displaystyle = \lim \limits_{x \to 0} \Bigg\{ \frac{(1+x)^6-1}{x} \times \frac{x}{(1+x)^2-1} \Bigg\}$

$\displaystyle = \lim \limits_{x \to 0} \Bigg\{ \frac{(1+x)^6-1^6}{(1+x) - 1} \times \frac{(1+x)-1}{(1+x)^2-1^2} \Bigg\}$

$\displaystyle \text{Let } y = 1+x$

$\displaystyle \text{When } x \rightarrow 0, \text{ then } 1+x \rightarrow 1 \Rightarrow y \rightarrow 1$

$\displaystyle = \lim \limits_{x \to 0} \Bigg\{ \frac{y^6-1^6}{y - 1} \times \frac{y-1}{y^2-1^2} \Bigg\}$

$\displaystyle = \frac{6}{2} ( 1^{6-2} )$

$\displaystyle = 3$

$\\$

$\displaystyle \text{Question 4: } \lim \limits_{x \to a} \frac{x^\frac{2}{7} - a^\frac{2}{7}}{x-a }$

Answer:

We have,

$\displaystyle \lim \limits_{x \to a} \frac{x^\frac{2}{7} - a^\frac{2}{7}}{x-a }$

$\displaystyle = \frac{2}{7}a^{\frac{2}{7}-1}$

$\displaystyle = \frac{2}{7} a^{\frac{-5}{7}}$

$\\$

$\displaystyle \text{Question 5: } \lim \limits_{x \to a} \frac{x^\frac{5}{7} - a^\frac{5}{7}}{x^\frac{2}{7} - a^\frac{2}{7} }$

Answer:

$\displaystyle \text{We have } \lim \limits_{x \to a} \frac{x^\frac{5}{7} - a^\frac{5}{7}}{x^\frac{2}{7} - a^\frac{2}{7} }$

$\displaystyle = \Bigg\{ \lim \limits_{x \to a} \frac{x^\frac{5}{7} - a^\frac{5}{7}}{x-a} \times \frac{x-a}{x^\frac{2}{7} - a^\frac{2}{7} } \Bigg\}$

$\displaystyle = \Bigg\{ \lim \limits_{x \to a} \frac{x^\frac{5}{7} - a^\frac{5}{7}}{x-a} \div \frac{x^\frac{2}{7} - a^\frac{2}{7} }{x-a} \Bigg\}$

$\displaystyle = \frac{5}{7} a^{\frac{5}{7}-1} \div \frac{2}{7} a^{\frac{2}{7}-1}$

$\displaystyle = \frac{5}{7} a^{\frac{-2}{7}} \div \frac{2}{7} a^{\frac{-5}{7}}$

$\displaystyle = \frac{5}{2} a^{\frac{-2}{7} +\frac{-5}{7} }$

$\displaystyle = \frac{5}{2} a^\frac{3}{7}$

$\\$

$\displaystyle \text{Question 6: } \lim \limits_{x \to -\frac{1}{2}} \frac{8x^3+1}{2x+1 }$

Answer:

$\displaystyle \text{We have } \lim \limits_{x \to -\frac{1}{2}} \frac{8x^3+1}{2x+1 }$

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

$\displaystyle \text{Let } y = 2x$

$\displaystyle \text{When } x \rightarrow -\frac{1}{2}, \text{ then } 2x \rightarrow -1$

$\displaystyle = \lim \limits_{x \to -\frac{1}{2}} \frac{y^3-(-1)^3}{y-(-1) }$

$\displaystyle = 3 (-1)^{3-1}$

$\displaystyle = 3 (-1)^2$

$\displaystyle = 3$

$\\$

$\displaystyle \text{Question 7: } \lim \limits_{x \to 27} \frac{(x^\frac{1}{3}+3)(x^\frac{1}{3}-3)}{ x-27}$

Answer:

$\displaystyle \text{We have, } \lim \limits_{x \to 27} \frac{(x^\frac{1}{3}+3)(x^\frac{1}{3}-3)}{ x-27}$

$\displaystyle = \lim \limits_{x \to 27} \frac{((x^\frac{1}{3})^2-3^2)}{ (x^\frac{1}{3})^3-3^3}$

$\displaystyle \text{When } x \rightarrow 27, \text{ then } x^\frac{1}{3} \rightarrow 3$

$\displaystyle \text{Let } y = x^\frac{1}{3}$

$\displaystyle = \lim \limits_{y\to 3} \frac{(y^2-3^2)}{ (y^3-3^3}$

$\displaystyle = \frac{2}{3} (3)^{2-3}$

$\displaystyle = \frac{2}{9}$

$\\$

$\displaystyle \text{Question 8: } \lim \limits_{x \to 4} \frac{x^3-64}{x^2-16 }$

Answer:

$\displaystyle \text{We have, } \lim \limits_{x \to 4} \frac{x^3-64}{x^2-16 }$

$\displaystyle = \lim \limits_{x \to 4} \frac{x^3-4^3}{x^2-4^2 }$

$\displaystyle = \lim \limits_{x \to 4} \Bigg\{ \frac{x^3-4^3}{x-4} \times \frac{x-4}{x^2-4^2} \Bigg\}$

$\displaystyle = \lim \limits_{x \to 4} \Bigg\{ \frac{x^3-4^3}{x-4} \div \frac{x^2-4^2}{x-4} \Bigg\}$

$\displaystyle = 3(4)^{3-1} \div 2(4)^{2-1}$

$\displaystyle = \frac{3 \times 16}{2 \times 4}$

$\displaystyle = 6$

$\\$

$\displaystyle \text{Question 9: } \lim \limits_{x \to 1} \frac{x^{15}-1}{ x^{10}-1}$

Answer:

$\displaystyle \text{We have, } \lim \limits_{x \to 1} \frac{x^{15}-1}{ x^{10}-1}$

$\displaystyle = \lim \limits_{x \to 1} \Bigg\{ \frac{x^{15}-1^{15}}{x-1} \times \frac{x-1}{x^{10}-1^{10}} \Bigg\}$

$\displaystyle = \lim \limits_{x \to 1} \Bigg\{ \frac{x^{15}-1^{15}}{x-1} \div \frac{x^{10}-1^{10}}{x-1} \Bigg\}$

$\displaystyle = 15(1)^{15-1} \div 10(1)^{10-1}$

$\displaystyle = \frac{15}{10}$

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

$\\$

$\displaystyle \text{Question 10: } \lim \limits_{x \to -1} \frac{x^3+1}{x+1 }$

Answer:

$\displaystyle \text{We have, } \lim \limits_{x \to -1} \frac{x^3+1}{x+1 }$

$\displaystyle = \lim \limits_{x \to -1} \frac{x^3-(-1)^3}{x-(-1) }$

$\displaystyle = 3(-1)^{3-1}$

$\displaystyle = 3$

$\\$

$\displaystyle \text{Question 11: } \lim \limits_{x \to a} \frac{x^\frac{2}{3}- a^\frac{2}{3}}{ x^\frac{3}{4}- a^\frac{3}{4}}$

Answer:

$\displaystyle \text{We have: } \lim \limits_{x \to a} \frac{x^\frac{2}{3}- a^\frac{2}{3}}{ x^\frac{3}{4}- a^\frac{3}{4}}$

$\displaystyle= \lim \limits_{x \to a} \frac{x^\frac{2}{3}- a^\frac{2}{3}}{x-a } \times \frac{x-a} { x^\frac{3}{4}- a^\frac{3}{4}}$

$\displaystyle= \lim \limits_{x \to a} \frac{x^\frac{2}{3}- a^\frac{2}{3}}{x-a } \div \frac{ x^\frac{3}{4}- a^\frac{3}{4}}{x-a}$

$\displaystyle= = \frac{2}{3} a^{\frac{2}{3}-1} \div \frac{3}{4} a^{\frac{3}{4}-1}$

$\displaystyle= = \frac{8}{9} a^{(-\frac{1}{3}+\frac{1}{4})}$

$\displaystyle= = \frac{8}{9} a^{-\frac{1}{12}}$

$\\$

$\displaystyle \text{Question 12: If } \lim \limits_{x \to 3} \frac{x^n-3^n}{x-3 } = 108, \text{ find the value of } n.$

Answer:

$\displaystyle \text{Given } \lim \limits_{x \to 3} \frac{x^n-3^n}{x-3 } = 108$

$\displaystyle\Rightarrow n(3)^{n-1} = 108$

$\displaystyle\Rightarrow n(3)^{n-1} = 4(3)^{3}$

$\displaystyle\Rightarrow n(3)^{n-1} = 4(3)^{4-1}$

$\displaystyle\Rightarrow n=4$

$\\$

$\displaystyle \text{Question 13: If } \lim \limits_{x \to a} \frac{x^9-a^9}{x-a } = 9, \text{ find the value of } a.$

Answer:

$\displaystyle \text{Given } \lim \limits_{x \to a} \frac{x^9-a^9}{x-a } = 9, \text{ find the value of } a.$

$\displaystyle\Rightarrow 9(a)^{9-1} = 9$

$\displaystyle\Rightarrow 9a^8 = 9$

$\displaystyle\Rightarrow a^8 = 1$

$\displaystyle\Rightarrow a = \pm 1$

$\\$

$\displaystyle \text{Question 14: If } \lim \limits_{x \to a} \frac{x^5-3^5}{x-a } = 405, \text{ find the value of } a.$

Answer:

$\displaystyle \text{Given } \lim \limits_{x \to a} \frac{x^5-3^5}{x-a } = 405$

$\displaystyle\Rightarrow 5(a)^{5-1} = 405$

$\displaystyle\Rightarrow 5a^4 = 405$

$\displaystyle\Rightarrow a^4 = 81$

$\displaystyle\Rightarrow a = \pm 3$

$\\$

$\displaystyle \text{Question 15: If } \lim \limits_{x \to a} \frac{x^9-a^9}{x-a } = \lim \limits_{x \to 5} (4+x), \text{ find the value of } a.$

Answer:

$\displaystyle \text{Given } \lim \limits_{x \to a} \frac{x^9-a^9}{x-a } = \lim \limits_{x \to 5} (4+x)$

$\displaystyle\Rightarrow 9(a)^{9-1} = 9$

$\displaystyle\Rightarrow a^8 = 1$

$\displaystyle\Rightarrow a = \pm 1$

$\\$

$\displaystyle \text{Question 16: If } \lim \limits_{x \to a} \frac{x^3-a^3}{x-a } = \lim \limits_{x \to 1} \frac{x^4-1}{x-1} , \text{ find the value of } a.$

Answer:

$\displaystyle \text{Given } \lim \limits_{x \to a} \frac{x^3-a^3}{x-a } = \lim \limits_{x \to 1} \frac{x^4-1}{x-1}$

$\displaystyle\Rightarrow 3(a)^{3-1} = 4(1)^{4-1}$

$\displaystyle\Rightarrow 3a^2 = 4$

$\displaystyle\Rightarrow a^2 = \frac{4}{3}$

$\displaystyle\Rightarrow a = \pm \frac{2}{\sqrt{3}}$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

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Class 11: Limits – Exercise 29.5

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