Evaluate the following limits: _ x 1 [ 2^ 2 x-2 -2^x+1 ^2(x-1) ]

Question

Evaluate the following limits: $\lim _{x \rightarrow 1}\left[\frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}\right]$

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Answer

$\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}$
$=\lim _{x \rightarrow 1} \frac{2^{2(x-1)}-2^x+1}{\sin ^2(x-1)}$
$\text { Put } x=1+\mathrm{h}, \therefore x-1=\mathrm{h}$
$\text { As } x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$\lim _{x \rightarrow 1} \frac{2^{2 x-2}-2^x+1}{\sin ^2(x-1)}=\lim _{h \rightarrow 0} \frac{2^{2 h}-2^{1+h}+1}{\sin ^2 h}$
$=\lim _{\mathrm{h} \rightarrow 0} \frac{2^{2 \mathrm{~h}}-2.2^{\mathrm{h}}+1}{\sin ^2 \mathrm{~h}}$
$=\lim _{h \rightarrow 0} \frac{\left(2^h-1\right)^2}{\sin ^2 h}$
$=\lim _{h \rightarrow 0}\left[\frac{\frac{\left(2^h-1\right)^2}{h^2}}{\frac{\sin ^2 h}{h^2}}\right] \ldots[\because h \rightarrow 0, h \neq 0]$
$=\frac{\left(\lim _{h \rightarrow 0} \frac{2^h-1}{h}\right)^2}{\left(\lim _{h \rightarrow 0} \frac{\sin h}{h}\right)^2}$
$=\frac{(\log 2)^2}{(1)^2}$
$\cdots\left[\because \lim _{x \rightarrow 0} \frac{\mathrm{a}^x-1}{x}=\log \mathrm{a}\right]$
$=(\log 2)^2$

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