Evaluate the following limit: _ x 0 8x 2x

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\tan8\text{x}}{\sin2\text{x}}$

✓

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\tan8\text{x}}{\sin2\text{x}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}{\tan8\text{x}}{}}{\lim\limits_{\text{x}\rightarrow0}{\sin2\text{x}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\frac{\tan8\text{x}}{8\text{x}}\times8\text{x}}{\lim\limits_{\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}\times2\text{x}}$ $\big[\because\sin3\text{x}=3\sin\text{x}-4\sin^3\text{x}\big]$
$=\frac{\lim\limits_{8\text{x}\rightarrow0}\frac{\tan8\text{x}}{8\text{x}}\times\frac{8\text{x}}{2\text{x}}}{\lim\limits_{2\text{x}\rightarrow0}\frac{\sin2\text{x}}{2\text{x}}}$ $\begin{bmatrix}\because\text{x}\rightarrow 0 \\8\text{x}\rightarrow0 \\2\text{x}\rightarrow0 \end{bmatrix}$
$=\frac{1\times\frac82}{1}$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\text{ and }\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac82$
$=4$

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