MHT CET · Maths · Limits
\(\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}=\)
- A \(\frac{1}{2}(\log 2)^2\)
- B \((\log 2)^2\)
- C \(2 \log 2\)
- D \(2(\log 2)^2\)
Answer & Solution
Correct Answer
(B) \((\log 2)^2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} & \lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}=\lim _{x \rightarrow 1} \frac{\left(2^{x-1}-1\right)^2}{\sin ^2(x-1)}=\lim _{x \rightarrow 1} \frac{\frac{\left(2^{x-1}-1\right)^2}{(x-1)^2}}{\frac{\sin ^2(x-1)}{(x-1)^2}} \\ & =\frac{(\log 2)^2}{1^2}=(\log 2)^2\end{aligned}\)
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