Differentiation Question 149
Question: If $ f(x)=\frac{\sin ({e^{x-2}}-1)}{in(x-1)}, $ then $ \underset{x\to 2}{\mathop{\lim }}f(x) $ is equal to
Options:
A) $ -2 $
B) $ -1 $
C) $ 0 $
D) 1
Show Answer
Answer:
Correct Answer: D
Solution:
[d] $ f(x)=\frac{\sin ( {e^{x-2}}-1 )}{In(x-1)} $ $ \underset{x\to 2}{\mathop{\lim }}\frac{\sin ( {e^{x-2}}-1 )}{In(x-1)}=L $ It is $ \frac{0}{0} $ (undefined) condition so using Hospital’s rule
$ \Rightarrow L=\underset{x\to 2}{\mathop{\lim }}[ \frac{{sin( {e^{x-2}}-1 )}}{{In(x-1)}} ] $
$ \Rightarrow L=\underset{x\to 2}{\mathop{\lim }}\frac{\cos ({e^{x-2}}-1).{e^{(x-2)}}}{1/(x-1)} $
$ \Rightarrow L=\underset{x\to 2}{\mathop{\lim }}\cos ( {e^{2-2}}-1 ){e^{2-2}}.(2-1) $
$ \Rightarrow L=\cos (0)e^{0}.1\Rightarrow L=1 $
BETA
