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Question

Question: \(\lim_{x \rightarrow 0}\frac{xe^{x} - \log(1 + x)}{x^{2}}\) equals...

lim⁡x→0xex−log⁡(1+x)x2\lim_{x \rightarrow 0}\frac{xe^{x} - \log(1 + x)}{x^{2}} equals

A

23\frac{2}{3}

B

13\frac{1}{3}

C

12\frac{1}{2}

D

32\frac{3}{2}

Answer

32\frac{3}{2}

Explanation

Solution

Let y=lim⁡x→0xex−log⁡(1+x)x2y = \lim_{x \rightarrow 0}\frac{xe^{x} - \log(1 + x)}{x^{2}} (00form)\left( \frac{0}{0}\text{form} \right)

Applying L–Hospital's rule, y=lim⁡x→0ex+xex−11+x2xy = \lim_{x \rightarrow 0}\frac{e^{x} + xe^{x} - \frac{1}{1 + x}}{2x} (00form)\left( \frac{0}{0}\text{form} \right)

y =lim⁡x→012[ex+ex+xex+1(1+x)2]=lim⁡x→012[1+1+0+1]=32= \lim_{x \rightarrow 0}\frac{1}{2}\left\lbrack e^{x} + e^{x} + xe^{x} + \frac{1}{(1 + x)^{2}} \right\rbrack = \lim_{x \rightarrow 0}\frac{1}{2}\lbrack 1 + 1 + 0 + 1\rbrack = \frac{3}{2}