Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{5\text{x}}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{5\text{x}}$
$=\frac{1}{5}\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}\times3$
$=\frac35\lim\limits_{\text{x}\rightarrow0}\frac{\sin3\text{x}}{3\text{x}}$
$=\frac{3}{5}\times1$ $\Big[\because\ \lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac{3}{5}$

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