Skip to content
Related Articles

Related Articles

Improve Article
Save Article
Like Article

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

  • Last Updated : 03 Mar, 2021

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}}

Attention reader! All those who say programming isn't for kids, just haven't met the right mentors yet. Join the  Demo Class for First Step to Coding Coursespecifically designed for students of class 8 to 12. 

The students will get to learn more about the world of programming in these free classes which will definitely help them in making a wise career choice in the future.

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}




My Personal Notes arrow_drop_up
Recommended Articles
Page :