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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}}

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}

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