limx→0 ex+sin⁡x−1log⁡(1+x)=

for the given limit dividing both N.R and D.R by x then we get =limx→0(ex-1+sinxxlog(1+x)x) =lim x→0(ex-1x)+limx→0(sinxx)log(limx→0(1+x)1x) =1+1loge =21 =2

limx→0 ex+sin⁡x−1log⁡(1+x)=

for the given limit dividing both N.R and D.R by x then we get =limx→0(ex-1+sinxxlog(1+x)x) =lim x→0(ex-1x)+limx→0(sinxx)log(limx→0(1+x)1x) =1+1loge =21 =2

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$lim_{x \to 0} \frac{e^{x} + \sin x - 1}{\log (1 + x)} =$

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$\frac{1}{3}$

$\frac{2}{3}$

answer is D.

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Detailed Solution

$f o r t h e g i v e n l i m i t d i v i d i n g b o t h N . R a n d D . R b y x t h e n w e g e t = l i m_{x \to 0} (\frac{\frac{e^{x} - 1 + \sin x}{x}}{\frac{\log (1 + x)}{x}}) = l i m \frac{_{x \to 0} (\frac{e^{x} - 1}{x}) + l i m_{x \to 0} (\frac{\sin x}{x})}{\log (l i m_{x \to 0} {(1 + x)}^{\frac{1}{x}})} = \frac{1 + 1}{\log e} = \frac{2}{1} = 2$