Question
Evaluate the following limit: $\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}\cos\text{x}+\sin\text{x}}{3\text{x}^2+\tan\text{x}}$
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Answer
$\lim\limits_{\text{x}\rightarrow0}\frac{5\text{x}\cos\text{x}+\sin\text{x}}{3\text{x}^2+\tan\text{x}}$ $=\lim\limits_{\text{x}\rightarrow0}\frac{5\cos\text{x}+3\frac{\sin\text{x}}{\text{x}}}{3\text{x}+\frac{\tan\text{x}}{\text{x}}}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}5\cos\text{x}\lim\limits_{\text{x}\rightarrow0}\frac{3\sin\text{x}}{\text{x}}}{\lim\limits_{\text{x}\rightarrow0}3\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$ $=\frac{\lim\limits_{\text{x}\rightarrow0}5\cos\text{x}+3\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}}{3\lim\limits_{\text{x}\rightarrow0}\text{x}+\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}}$ $=\frac{5\times\cos0+3\times1}{3\times0+1}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1,\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$ $=\frac{5+3}{1}$ $=8$
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