Functions Question 194
Question: If $ f(x)=\frac{\sin ({e^{x-2}}-1)}{\log (x-1)}, $ then $ \underset{x\to 2}{\mathop{\lim }},f(x) $ is given by
Options:
A) ?2
B) ?1
C) 0
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
$ \underset{x\to 2}{\mathop{\lim }},f(x)=\underset{x\to 2}{\mathop{\lim }}\frac{\sin ,({e^{x-2}}-1)}{\log ,(x-1)} $ $ =\underset{t\to 0}{\mathop{\lim }},\frac{\sin ,(e^{t}-1)}{\log ,(1+t)} $ , {Putting $ x=2+t} $ $ =\underset{t\to 0}{\mathop{\lim }},\frac{\sin ,(e^{t}-1)}{e^{t}-1}.\frac{e^{t}-1}{t}.\frac{t}{\log ,(1+t)} $ $ =\underset{t\to 0}{\mathop{\lim }},\frac{\sin ,(e^{t}-1)}{e^{t}-1}.( \frac{1}{1!}+\frac{t}{2!}+… )\times [ \frac{1}{( 1-\frac{1}{2}t+\frac{1}{3}t^{2}-… )} ] $ $ =1.1.1=1,(\because Ast\to 0,e^{t}-1\to 0). $
BETA
