MCQ
$\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}=$
  • A
    $2 \log 2$
  • B
    $2(\log 2)^2$
  • C
    $\frac{1}{2}(\log 2)^2$
  • $(\log 2)^2$

Answer

Correct option: D.
$(\log 2)^2$
(d) : We have, $\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}$
Given limit is of the form $\frac{0}{0}$. So, applying L' Hopital's rule, we get
$
\lim _{x \rightarrow 1} \frac{2^{2 x-2} \log 2(2)-2^x \log 2+0}{2 \sin (x-1) \cos (x-1)}
$
$
=\lim _{x \rightarrow 1} \frac{2^{2 x-2} 2 \times(\log 2)-2^x \log 2}{\sin 2(x-1)}
$ $\left(\frac{0}{0} \text { form }\right)
$
Again applying L'Hopital's Rule,
$
\lim _{x \rightarrow 1} \frac{2 \log 2 \times 2^{2 x-2}(2 \log 2)-2^x(\log 2)^2}{2 \cos 2(x-1)}
$
$
=\frac{4(\log 2)^2-2(\log 2)^2}{2}=\frac{2(\log 2)^2}{2}=(\log 2)^2
$

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