Question

Evaluate:
$\rm \displaystyle \lim_{x\to 0}\ \frac{2\sin x^{\circ} -\sin 2x^{\circ}}{x^{3}}$

Answer 1

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Answer 2

Answer:

1

Step-by-step explanation:

We have to evaluate,

$\lim_{x \to \ 0} \frac{2Sinx-Sin2x}{x^{3} }$

=$\lim_{x \to \ 0} \frac{2Sinx(1-Cosx)}{x^{3} }$

=$\lim_{x \to \ 0} \frac{2Sinx.2Sin^{2}(\frac{x}{2} ) }{x^{3} }$ {Since $Cos2x=1-2Sin^{2}x$}

=$4\lim_{x \to \ 0} \frac{Sinx}{x}. \lim_{x \to \ 0} (\frac{Sin\frac{x}{2} }{\frac{x}{2} } )^{2}.\frac{1}{4}$ {Here we will use the formula, $\lim_{x \to \ 0} \frac{Sinx}{x}=1$}

=$\frac{4}{4}.1. [ \lim_{\frac{x}{2}  \to \ 0} (\frac{Sin\frac{x}{2} }{\frac{x}{2} } ) ]^{2}$ {Since x tends to 0, so, $\frac{x}{2}$ also tends to 0}

=1×1×1²

=1 (Answer)