Evaluate the following limits:

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

Answer:

\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}

Answer:

\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}

Answer:

\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}

Answer:

\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}

Answer:

\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}

Answer:

\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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