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Limits — Maths STD 11 Science — Question

Maharashtra BoardEnglish MediumSTD 11 ScienceMathsLimits2 Marks
Question
Evaluate the following limits:
$\lim _{x \rightarrow 1}\left[\frac{4^{x-1}-2^x+1}{(x-1)^2}\right]$
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Answer

$ \lim _{x \rightarrow 1}\left[\frac{4^{x-1}-2^x+1}{(x-1)^2}\right]$
$\text { put } x-1=\mathbf{h}$
$\therefore \quad x=1+\mathrm{h}$
$\text { As } x \rightarrow 1, \mathrm{~h} \rightarrow 0$
$\text { Required limit }$
$=\lim _{h \rightarrow 0} \frac{4^h-2^{1+h}+1}{h^2}$
$=\lim _{h \rightarrow 0} \frac{\left(2^2\right)^h-2^1 \cdot 2^h+1}{h^2}$
$=\lim _{h \rightarrow 0} \frac{\left(2^h\right)^2-2\left(2^h\right)+1}{h^2}$
$=\lim _{h \rightarrow 0} \frac{\left(2^h-1\right)^2}{h^2}$
$=\lim _{h \rightarrow 0}\left(\frac{2^h-1}{h}\right)^2$
$=(\log 2)^2 \quad \ldots\left[\because \lim _{x \rightarrow 0} \frac{a^x-1}{x}=\log a\right]$

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