MCQ
$\lim _{x \rightarrow 0}\left[\frac{1}{x}-\frac{\log (1+x)}{x^2}\right]=$
  • $\frac{1}{2}$
  • B
    $\frac{-1}{2}$
  • C
    1
  • D
    -1

Answer

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

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