Limits Continuity And Differentiability Question 89
Question: Let $ f(x)=\frac{{{(e^{x}-1)}^{2}}}{\sin ( \frac{x}{a} )\log ( 1+\frac{x}{4} )} $ for $ x\ne 0, $ and $ f(0)=12 $ . If f is continuous at $ x=0 $ , then the value of a is equal to
Options:
A) 1
B) -1
C) 2
D) 3
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to 0}{\mathop{Lt}}\frac{{{(e^{x}-1)}^{2}}}{\sin ( \frac{x}{a} )\log ( 1+\frac{x}{4} )} $
$ =\underset{x\to 0}{\mathop{Lt}}\frac{\frac{{{(e^{x}-1)}^{2}}}{x}.x^{2}}{\frac{x}{a}.\frac{\sin ( \frac{x}{a} )}{( \frac{x}{a} )}.\frac{\log ( 1+\frac{x}{4} )}{\frac{x}{4}}.\frac{x}{4}}\Rightarrow 4a=12 $
$ \Rightarrow a=3 $
BETA
