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Class 11 RD Sharma Solutions – Chapter 29 Limits – Exercise 29.5 | Set 2

Last Updated : 03 Mar, 2021

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Question 11. $\lim_{x \to a}\frac {x^{\frac 2 3}-a^{\frac 2 3}} {x^{\frac 3 4}-a^{\frac 3 4}}$

Solution:

$\lim_{x \to a}\frac {x^{\frac 2 3}-a^{\frac 2 3}} {x^{\frac 3 4}-a^{\frac 3 4}}$

Dividing numerator and denominator by x-a

$=\lim_{x \to a}\frac {\frac {x^{\frac 2 3}-a^{\frac 2 3}} {x-a}} {\frac {x^{\frac 3 4}-a^{\frac 3 4}} {x-a}}$

$=\frac {\lim_{x \to a}\frac {x^{\frac 2 3}-a^{\frac 2 3}} {x-a}} {\lim_{x \to a}\frac {x^{\frac 3 4}-a^{\frac 3 4}} {x-a}}$

Applying the formula $\lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}$ here, $n=\frac 2 3$ for numerator and $n=\frac 3 4$ for denominator

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

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

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

Question 12. If $\lim_{x \to 3}\frac {x^n-3^n} {x-3}=108$ , find the value of n.

Solution:

$\lim_{x \to 3}\frac {x^n-3^n} {x-3}=108$

LHS = $\lim_{x \to 3}\frac {x^n-3^n} {x-3}$

Applying the formula $\lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}$

LHS = $n(3)^{n-1}$

RHS = 108

LHS = RHS

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

$\implies n(3)^{n-1}=2\times2\times3\times3\times3$

$\implies n(3)^{n-1}=2^2\times3^3$

$\implies n(3)^{n-1}=4\times3^{4-1}$

$\implies n=4$

Question 13. If $\lim_{x \to a}\frac {x^9-a^9} {x-a}=9$ , find all possible values of a.

Solution:

$\lim_{x \to a}\frac {x^9-a^9} {x-a}=9$

LHS= $\lim_{x \to a}\frac {x^9-a^9} {x-a}$

Applying the formula $\lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}$

LHS = $9(a)^{9-1}$

RHS = 9

LHS = RHS

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

$\implies 9(a)^{8}=9$

$\implies (a)^{8}=\frac 9 9$

$\implies a^8=1$

$\implies a= \pm1$

$\implies$ a=1 and a=-1

Question 14. If $\lim_{x \to a}\frac {x^5-a^5} {x-a}=405$ , find all possible values of a.

Solution:

$\lim_{x \to a}\frac {x^5-a^5} {x-a}=405$

LHS = $\lim_{x \to a}\frac {x^5-a^5} {x-a}$

Applying the formula $\lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}$

LHS = $5a^{5-1}$

RHS = 405

LHS = RHS

$\implies 5a^{5-1}=405$

$\implies 5a^4=405$

$\implies a^4=\frac {405} 5$

$\implies a^4=81$

$\implies a=\pm 3$

$\implies$ a=3 and a=-3

Question 15. If $\lim_{x \to a}\frac {x^9-a^9} {x-a}=\lim_{x \to 5}(4+x)$ , find all possible values of a.

Solution:

$\lim_{x \to a}\frac {x^9-a^9} {x-a}=\lim_{x \to 5}(4+x)$

LHS= $\lim_{x \to a}\frac {x^9-a^9} {x-a}$

Applying the formula $\lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}$

LHS= $9a^{9-1}=9a^8$

RHS= $\lim_{x \to 5}(4+x)=4+5=9$

LHS=RHS

$\implies 9a^8=9$

$\implies a^8=1$

$\implies a=\pm1$

$\implies$ a=1 and a=-1

Question 16. If $\lim_{x \to a}\frac {x^3-a^3} {x-a}=\lim_{x \to 1}\frac {x^4-1}{x-1}$ , find all possible values of a.

Solution:

$\lim_{x \to a}\frac {x^3-a^3} {x-a}=\lim_{x \to 1}\frac {x^4-1}{x-1}$

LHS= $\lim_{x \to a}\frac {x^3-a^3} {x-a}$

Applying the formula $\lim_{x \to a} \frac {x^{n}-a^{n}} {x-a}=na^{n-1}$

LHS= $3a^{3-1}=3a^2$

RHS= $4(1)^{4-1}=4$

LHS=RHS

$\implies 3a^2=4$

$\implies a^2=\frac 4 3$

$\implies a= \pm \frac 2 {\sqrt3}$

$\implies a=\frac 2 {\sqrt3}$ and $a=-\frac 2 {\sqrt3}$