Evaluate the following limits:

$\displaystyle \text{ Question 1: } \lim \limits_{x \to \pi } \ \frac{1+\cos x}{\tan^2 x}$

$\displaystyle \lim \limits_{x \to \pi } \ \frac{1+\cos x }{\tan^2 x}$

$\displaystyle =\lim \limits_{x \to \pi } \ \frac{1+\cos x }{\sin^2 x} \times \cos^2 x$

$\displaystyle =\lim \limits_{x \to \pi } \ \frac{1+\cos x }{1 - \cos^2 x} \times \cos^2 x$

$\displaystyle =\lim \limits_{x \to \pi } \ \frac{1+\cos x }{(1 - \cos x)(1+ \cos x)} \times \cos^2 x$

$\displaystyle =\lim \limits_{x \to \pi } \ \frac{\cos^2 x }{(1 - \cos x)}$

$\displaystyle = \frac{\cos^2 \pi }{(1 - \cos \pi)} = \frac{(-1)^2}{1-(-1)} = \frac{1}{2}$

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$\displaystyle \text{ Question 2: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\mathrm{cosec}^2 x - 2}{\cot x - 1}$

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\mathrm{cosec}^2 x - 2}{\cot x - 1}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1+\cot^2 x - 2}{\cot x - 1}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{\cot^2 x - 1}{\cot x - 1}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{(\cot x - 1)(\cot x + 1)}{(\cot x - 1)}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \cot x + 1$

$\displaystyle = \cot \frac{\pi}{4} + 1 = 2$

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$\displaystyle \text{ Question 3: } \lim \limits_{x \to \frac{\pi}{6} } \ \frac{\cot^2 x - 3}{\mathrm{cosec} x - 2}$

$\displaystyle \lim \limits_{x \to \frac{\pi}{6} } \ \frac{\cot^2 x - 3}{\mathrm{cosec} - 2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{6} } \ \frac{\mathrm{cosec}^2 x - 1 - 3}{\mathrm{cosec} - 2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{6} } \ \frac{\mathrm{cosec}^2 x - 4}{\mathrm{cosec} - 2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{6} } \ \frac{(\mathrm{cosec} x - 2)(\mathrm{cosec} x + 2)}{\mathrm{cosec} - 2}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{6} } \ \mathrm{cosec} x + 2$

$\displaystyle = 4$

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$\displaystyle \text{ Question 4: } \lim \limits_{x \to \frac{\pi}{4} } \ \frac{2 - \mathrm{cosec}^2 x}{1 - \cot x}$

$\displaystyle \lim \limits_{x \to \frac{\pi}{4} } \ \frac{2 - \mathrm{cosec}^2 x}{1 - \cot x}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{2 - 1-\cot^2 x}{1 - \cot x}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{1-\cot^2 x}{1 - \cot x}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ \frac{(1-\cot x)(1+ \cot x)}{1 - \cot x}$

$\displaystyle = \lim \limits_{x \to \frac{\pi}{4} } \ 1+ \cot x$

$\displaystyle = 1+ \cot \frac{\pi}{4} = 1+1 = 2$

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$\displaystyle \text{ Question 5: } \lim \limits_{x \to \pi } \ \frac{\sqrt{2 + \cos x}-1}{(\pi - x)^2}$

$\displaystyle \lim \limits_{x \to \pi } \ \frac{\sqrt{2+\cos x}-1}{(\pi - x)^2}$

$\displaystyle = \lim \limits_{x \to \pi } \ \frac{(\sqrt{2+\cos x}-1)(\sqrt{2+\cos x}+1)}{(\pi - x)^2 (\sqrt{2+\cos x}+1)}$

$\displaystyle = \lim \limits_{x \to \pi } \ \frac{ 2 + \cos x - 1 }{(\pi - x)^2 (\sqrt{2+\cos x}+1)}$

$\displaystyle = \lim \limits_{x \to \pi } \ \frac{ 1 + \cos x }{(\pi - x)^2 (\sqrt{2+\cos x}+1)}$

$\displaystyle \text{Let } x = \pi - h, \text{ when } x \rightarrow \pi, \text{ then } h \rightarrow 0$

$\displaystyle = \lim \limits_{h \to 0 } \ \frac{ 1 + \cos (\pi - h) }{(\pi - (\pi - h))^2 (\sqrt{2+\cos (\pi - h)}+1)}$

$\displaystyle = \lim \limits_{h \to 0 } \frac{1 - \cos h}{h^2( \sqrt{2 - \cos h} +1) }$

$\displaystyle = \lim \limits_{h \to 0 } \Bigg[ \frac{2 \sin^2 \frac{h}{2} }{4 \times \frac{h^2}{4} ( \sqrt{2 - \cos h} +1) } \Bigg]$

$\displaystyle = \lim \limits_{h \to 0 } \Bigg[ \frac{1}{2} \Bigg( \frac{\sin \frac{h}{2}}{\frac{h}{2}} \Bigg)^2 \times \frac{1}{\sqrt{2 - \cos h} +1} \Bigg]$

$\displaystyle = \frac{1}{2} \times 1 \times \frac{1}{\sqrt{2 - \cos 0} +1} = \frac{1}{4}$

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$\displaystyle \text{ Question 6: } \lim \limits_{x \to \frac{3\pi}{4} } \ \frac{1 + \mathrm{cosec}^3 x}{\cot^2 x}$

$\displaystyle \lim \limits_{x \to \frac{3\pi}{4} } \ \frac{1 + \mathrm{cosec}^3 x}{\cot^2 x}$

$\displaystyle = \lim \limits_{x \to \frac{3\pi}{4} } \ \frac{ (1 + \mathrm{cosec} x ) (1^2+ \mathrm{cosec}^2 x - \mathrm{cosec} x ) }{(\mathrm{cosec}^2 x -1)}$

$\displaystyle = \lim \limits_{x \to \frac{3\pi}{4} } \ \frac{ (1 + \mathrm{cosec} x ) (1^2+ \mathrm{cosec}^2 x - \mathrm{cosec} x ) }{(\mathrm{cosec} x -1)(\mathrm{cosec} x +1)}$

$\displaystyle = \lim \limits_{x \to \frac{3\pi}{4} } \ \frac{ (1^2+ \mathrm{cosec}^2 x - \mathrm{cosec} x ) }{(\mathrm{cosec} x -1)}$

$\displaystyle = \frac{ (1^2+ \mathrm{cosec}^2 \frac{3\pi}{4} - \mathrm{cosec} \frac{3\pi}{4} ) }{(\mathrm{cosec} \frac{3\pi}{4} -1)}$

$\displaystyle = \frac{1+1+1}{-1-1} = \frac{-3}{2}$

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