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Question: \(\lim_{x \rightarrow 0}\frac{\sin x + \log(1 - x)}{x^{2}}\) is equal to...

lim⁡x→0sin⁡x+log⁡(1−x)x2\lim_{x \rightarrow 0}\frac{\sin x + \log(1 - x)}{x^{2}} is equal to

A

0

B

12\frac{1}{2}

C

−12- \frac{1}{2}

D

None of these

Answer

−12- \frac{1}{2}

Explanation

Solution

Apply L- Hospital rule, we get,

lim⁡x→0cos⁡x−11−x2x=lim⁡x→0−sin⁡x−1(1−x)22=−12\lim_{x \rightarrow 0}\frac{\cos x - \frac{1}{1 - x}}{2x} = \lim_{x \rightarrow 0}\frac{- \sin x - \frac{1}{(1 - x)^{2}}}{2} = - \frac{1}{2}

Alternative method : lim⁡x→0sin⁡x+log⁡(1−x)x2\lim_{x \rightarrow 0}\frac{\sin x + \log(1 - x)}{x^{2}}

=lim⁡x→0(x−x33!+x55!−……)x2+lim⁡x→0(−x−x22−x33−x44−……)x2= \lim _ { x \rightarrow 0 } \frac { \left( x - \frac { x ^ { 3 } } { 3 ! } + \frac { x ^ { 5 } } { 5 ! } - \ldots \ldots \right) } { x ^ { 2 } } + \lim _ { x \rightarrow 0 } \frac { \left( - x - \frac { x ^ { 2 } } { 2 } - \frac { x ^ { 3 } } { 3 } - \frac { x ^ { 4 } } { 4 } - \ldots \ldots \right) } { x ^ { 2 } } (∵sin⁡x=x−x33!+x55!−......and log(1−x)=−x−x22−x33........)\left( \because\sin x = x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - ......\text{and log}(1 - x) = - x - \frac{x^{2}}{2} - \frac{x^{3}}{3}........ \right)Hence, lim⁡x→0−x22−x3(13!+13)−x44....x2=−12.\lim_{x \rightarrow 0}\frac{\frac{- x^{2}}{2} - x^{3}\left( \frac{1}{3!} + \frac{1}{3} \right) - \frac{x^{4}}{4}....}{x^{2}} = - \frac{1}{2}.