MCQ
$\lim _{x \rightarrow 0} \frac{x e^x-\log (1+x)}{x^2}$ equals
  • A
    $\frac{2}{3}$
  • B
    $\frac{1}{3}$
  • C
    $\frac{1}{2}$
  • $\frac{3}{2}$

Answer

Correct option: D.
$\frac{3}{2}$
(D)
Let $y=\lim _{x \rightarrow 0} \frac{x e ^x-\log (1+x)}{x^2}$
Applying L-Hospital's rule, we get
$y=\lim _{x \rightarrow 0} \frac{ e ^x+x e ^x-\frac{1}{1+x}}{2 x}$
$=\lim _{x \rightarrow 0} \frac{1}{2}\left[ e ^x+ e ^x+x e ^x+\frac{1}{(1+x)^2}\right]$
$=\frac{1}{2}(1+1+0+1)=\frac{3}{2}$

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