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 andn=\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} anda=-\frac 2 {\sqrt3}
