Functions Question 202
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{xe^{x}-\log (1+x)}{x^{2}} $ equals
[RPET 1996]
Options:
A) $ \frac{2}{3} $
B) $ \frac{1}{3} $
C) $ \frac{1}{2} $
D) $ \frac{3}{2} $
Show Answer
Answer:
Correct Answer: D
Solution:
Let $ y=\underset{x\to 0}{\mathop{\lim }},\frac{x,e^{x}-\log ,(1+x)}{x^{2}} $ , $ ( \frac{0}{0},form ) $ Applying L-Hospital’s rule, $ y=\underset{x\to 0}{\mathop{\lim }},\frac{e^{x}+x,e^{x}-\frac{1}{1+x}}{2x} $ , $ ( \frac{0}{0},form ) $ $ y=\underset{x\to 0}{\mathop{\lim }},\frac{1}{2},[ e^{x}+e^{x}+x,e^{x}+\frac{1}{{{(1+x)}^{2}}} ] $ $ y=\underset{x\to 0}{\mathop{\lim }},\frac{1}{2},[1+1+0+1]=\frac{3}{2} $ .
BETA
