Question

lim x=>0 3^sinx-1 / x
solve using a^x-1/x =loga formula​

Answer 1

$\huge\boxed{\underline{\underline{\green{Solution}}}} </p><p>$

$\displaystyle \lim_{x \to 0} \frac{ {3}^{sinx} - 1 }{x}$

$\displaystyle let \: \: \: y \: = \: sinx$

$so \: \: \: \: \displaystyle {y \to 0} \: \: as \: \: \: {x \to 0}$

Hence

$\displaystyle \lim_{x \to 0} \: \frac{ {3}^{sinx} - 1 }{x}$

=$\displaystyle \lim_{x \to 0} \: (\frac{ {3}^{sinx} - 1 }{sinx} \times \frac{sinx}{x} )$

$= \displaystyle \lim_{x \to 0} \: \frac{ {3}^{sinx} - 1 }{sinx} \times \lim_{x \to 0} \: \frac{sinx}{x}$

$= \displaystyle \lim_{y \to 0} \: \frac{ {3}^{y} - 1 }{y} \times \lim_{x \to 0} \: \frac{sinx}{x}$

$= \: log3 \times 1 \: \: \: ( \because \displaystyle \lim_{y \to 0} \: \frac{ {a}^{x} - 1 }{x} = log \: a \: \: \: and \: \: \displaystyle \lim_{x \to 0} \: \frac{ sinx}{x} = 1 \: )$

$= log \: 3$

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