Functions Question 156
Question: $ \underset{x\to 0}{\mathop{\lim }},\frac{\sin x+\log (1-x)}{x^{2}} $ is equal to
[Roorkee 1995]
Options:
A) 0
B) $ \frac{1}{2} $
C) $ -\frac{1}{2} $
D) None of these
Show Answer
Answer:
Correct Answer: C
Solution:
Apply L-Hospital?s rule, we get $ \underset{x\to 0}{\mathop{\lim }}\frac{\cos x-\frac{1}{1-x}}{2x}=\underset{x\to 0}{\mathop{\lim }}\frac{-\sin x-\frac{1}{{{(1-x)}^{2}}}}{2}=-\frac{1}{2} $ . Aliter : $ \underset{x\to 0}{\mathop{\lim }}\frac{\sin x+\log ,(1-x)}{x^{2}} $ $ =\underset{x\to 0}{\mathop{\lim }},\frac{( x-\frac{x^{3}}{3!}+\frac{x^{5}}{5!}-… )}{x^{2}}+\underset{x\to 0}{\mathop{\lim }},\frac{( -x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-\frac{x^{4}}{4}-… )}{x^{2}} $ $ ( \because \sin x=x-\frac{x^{3}}{3,!}+\frac{x^{5}}{5,!}-.. . $ and $ . \log ,(1-x)=-x-\frac{x^{2}}{2}-\frac{x^{3}}{3}-.. ) $ $ =\underset{x\to 0}{\mathop{\lim }}\frac{\frac{-x^{2}}{2}-x^{3}( \frac{1}{3!}+\frac{1}{3} )-\frac{x^{4}}{4}…}{x^{2}}=-\frac{1}{2}. $
BETA
