Evaluate the following limit: _ x 0 3x-2x 3x- ^2x - Vidyadip

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}-2\text{x}}{3\text{x}-\sin^2\text{x}}$

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Answer

$\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}-2\text{x}}{3\text{x}-\sin^2\text{x}}$
$=\lim\limits_{\text{x}\rightarrow0}\frac{\frac{\tan3\text{x}}{\text{x}}-\frac{\tan2\text{x}}{\text{x}}}{\frac{3\text{x}}{\text{x}}-\frac{\sin^2\text{x}}{\text{x}}}$
$=\frac{\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan3\text{x}}{3\text{x}}\times3\Big)-\Big(\lim\limits_{\text{x}\rightarrow0}\frac{\tan2\text{x}}{2\text{x}}\times2\Big)}{\Big(\lim\limits_{\text{x}\rightarrow0}\frac{3\text{x}}{2}\Big)-\lim\limits_{\text{x}\rightarrow0}\frac{(\sin\text{x})^2}{\text{x}}}$
$=\frac{3-2}{3-\big(\frac{\sin\text{x}}{\text{x}}\big)^2\times\text{x}}$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\tan\text{x}}{\text{x}}=1\Big]$
$=\frac{3-2}{3-0}=\frac13$ $\Big[\because\lim\limits_{\text{x}\rightarrow0}\frac{\sin\text{x}}{\text{x}}=1\Big]$
$=\frac13$

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