Evaluate the following limit: _ x 1+ ^2x

Question

Evaluate the following limit:
$\lim\limits_{\text{x}\rightarrow\pi}\frac{1+\cos\text{x}}{\tan^2\text{x}}$

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Answer

$\lim\limits_{\text{x}\rightarrow\pi}\frac{1+\cos\text{x}}{\tan^2\text{x}}$
As$\text{ x}\rightarrow\pi,\text{x}-\pi\rightarrow0,$let $\text{ x }-\pi=\text{y}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1+\cos(\pi+\text{y})}{\tan^2(\pi+\text{y})}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{1-\cos\text{y}}{\tan^2\text{y}}$
$=\lim\limits_{\text{y}\rightarrow0}\frac{2\sin^2\frac{\text{y}}{2}}{\tan^2\text{y}}$
$=\frac{\lim\limits_{\text{y}\rightarrow0}2\sin^2\frac{\text{y}}{2}}{{\lim\limits_{\text{y}\rightarrow0}\tan^2\text{y}}}$
$=\frac{2\Bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\frac{\sin\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{\text{y}^2}{4}}{\bigg(\lim\limits_{\text{y}\rightarrow0}\frac{\tan\text{y}}{\text{y}}\bigg)\times\text{y}^2}$
$=\frac{2\times1\times\frac{\text{y}^2}{4}}{1\times\text{y}^2}$ $\begin{bmatrix}\therefore\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1\\\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1 \end{bmatrix}$
$=2\times1\times\frac{1}{4}$
$=\frac{1}{2}$

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