Question

explain step by step​

Answer 1

$\begin{gathered}\Large{\bold{\pink{\underline{Formula \: Used \::}}}} \end{gathered}$

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$: \implies \: \tt \: \tt \:\lim_{x\to0} \: \dfrac{sinx}{x} = 1$

$: \implies \: \tt \: \tt \:\lim_{x\to0}\dfrac{ {a}^{x} - 1}{x} = log(a)$

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$\large\underline\purple{\bold{Solution :- }}$

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$: \implies \: \tt \: \:\lim_{x\to0}\dfrac{ {4}^{x} + {4}^{ - x} - 2 }{x \: sinx}$

☆ On substituting directly the value of x, we get indeterminant form

$= \tt \: \:\lim_{x\to0}\dfrac{ {4}^{x} + \dfrac{1}{ {4}^{x} } - 2}{x} \times \:\lim_{x\to0}\dfrac{1}{\dfrac{sinx}{x} \times x}$

$= \tt \: \:\lim_{x\to0}\dfrac{ {4}^{2x} + 1 - 2 \times {4}^{x} }{ {x}^{2} \times {4}^{x} } \times \:\lim_{x\to0}\dfrac{x}{sinx}$

$= \tt\:\lim_{x\to0}\dfrac{1}{ {4}^{x} } \times \:\lim_{x\to0} \dfrac{ {( {4}^{x} - 1)}^{2} }{ {x}^{2} } \times 1$

$= \tt \:1 \times \:\lim_{x\to0}\dfrac{ {4}^{x} - 1 }{x} \times \:\lim_{x\to0}\dfrac{ {4}^{x} - 1 }{x}$

$\tt \: = log(4) \times log(4)$

$\tt\implies \: = {( log(4)) }^{2}$

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