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
$\lim\limits_{\text{x}\rightarrow{{\pi}}}\frac{1+\cos\text{x}}{\tan^2\text{ x}}$ $\Rightarrow\text{x}\rightarrow{\pi},\text{x}-{\pi}\rightarrow0,$ let $\text{y}=\text{x}-\pi$ $\Rightarrow\lim\limits_{\text{x}-{\pi}\rightarrow{{0}}}\frac{1+\cos\text{x}}{\tan^2\text{x}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1+\cos(\pi+\text{y})}{\tan^2(\pi+\text{y})}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{1-\cos\text{y}}{\tan^2\text{y}}$ $=\lim\limits_{\text{y}\rightarrow{0}}\frac{2\sin^2\frac{\text{y}}{2}}{\tan^2\text{y}}$ $=2\Bigg(\lim\limits_{\text{y}\rightarrow{0}}\frac{\sin\frac{\text{y}}{2}}{\frac{\text{y}}{2}}\Bigg)^2\times\frac{\text{y}^2}{4}\times\frac{1}{\Big(\lim\limits_{\text{y}\rightarrow{0}}\frac{\tan\text{y}}{\text{y}}\Big)^2\times\text{y}^2}$ $=2\times1\times\frac{\text{y}^2}{4}\times\frac{1}{1\times\text{y}^2}$ $\Big[\because\lim\limits_{\theta\rightarrow0}\frac{\sin\theta}{\theta}=1,\lim\limits_{\theta\rightarrow0}\frac{\tan\theta}{\theta}=1\Big]$ $=\frac{1}{2}$
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