Question
Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\cos\text{x}+2\sin\text{x}}{\text{x}^2+\tan\text{x}}$

Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\text{x}\cos\text{x}+2\sin\text{x}}{\text{x}^2+\tan\text{x}}$
Dividing each term by x
$=\lim\limits_{\text{x}\rightarrow0}\frac{\cos\text{x}+\frac{2\sin\text{x}}{\text{x}}}{\text{x}+\frac{\tan\text{x}}{\text{x}}}$
$=\frac{\lim\limits_{\text{x}\rightarrow0}\cos\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{2\sin\text{x}}{\text{x}}}{\lim\limits_{\text{x}\rightarrow0}\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$
$=\frac{\cos0+2\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}{0+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$
$=\frac{1+2}{0+1}=3$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=3$

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