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Substitution Method to Remove Indeterminate Form

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Question

Evaluate : ${lim}_{x \to 0} \frac{sin 2 x}{5 x}$

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Solution

$\begin{matrix} W e h a v e {lim}_{x \to 0} \frac{sin (2 x)}{5 x} = \frac{1}{5} {lim}_{x \to 0} \frac{sin (2 x)}{x} N o w, b y u sin g L^{'} H o s p i t a l R u l e {lim}_{x \to 0} \frac{sin (2 x)}{x} = {lim}_{x \to 0} \frac{\frac{d}{d x} [sin 2 x]}{\frac{d (x)}{d x}} = \frac{1}{5} {lim}_{x \to 0} \frac{cos (2 x) \frac{d (2 x)}{d x}}{\frac{d (x)}{d x}} = \frac{1}{5} {lim}_{x \to 0} \frac{2. cos (2 x) \frac{d (x)}{d x}}{\frac{d (x)}{d x}} = \frac{1}{5} {lim}_{x \to 0} \frac{2 cos (2 x) .1}{\frac{d (x)}{d x}} (p o w e r r u l e) = \frac{1}{5} . \frac{cos (2 {lim}_{x \to 0} x)}{{lim}_{x \to 0} 1} = \frac{1}{5} . \frac{2 cos (2.0)}{1} = \frac{1}{5} . \frac{2 cos 0}{1} = \frac{1}{5} \times 2 = \frac{2}{5} H e n c e, i t i s t h e r e q u i r e d a n s w e r . \end{matrix}$

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