MCQ
$\lim _{x \rightarrow 0} \frac{2^x-1}{(1+x)^{1 / 2}-1}=$
  • A
    $\log 2$
  • $\log 4$
  • C
    $\log \sqrt{2}$
  • D
    $3 \log 2$

Answer

Correct option: B.
$\log 4$
(B)
Applying L-Hospital's rule, we get
$\lim _{x \rightarrow 0} \frac{2^x-1}{(1+x)^{1 / 2}-1}=\lim _{x \rightarrow 0} \frac{2^x \log 2}{\frac{1}{2}(1+x)^{-1 / 2}}$
$=2 \log 2=\log 4$

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