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Limits Continuity and Differentiability

Section: Mathematics

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Question : 7 of 61

Marks: +1, -0

\lim\limits_{x\to 0}\;\frac{(\csc x-\cot x)(e^{x}-e^{-x})}{\sqrt{3}-\sqrt{2+\cos x}}=

[AP EAPCET 23 May 2025 S2]

$\;3\sqrt{2}$
$\;2\sqrt{3}$
$\;3\sqrt{3}$
$4\sqrt{3}$

Solution:

\lim\limits_{x\to 0}\;\frac{(\csc x-\cot x)(e^{x}-e^{-x})}{\sqrt{3}-\sqrt{2+\cos x}}

Convert,

\csc x

and

\cot x

into

\sin x

and

e^{x}-e^{-x}=2\sinh x

\;=\lim\limits_{x\to 0}\;\frac{ \left( \frac{2\sin^{2}\frac{x}{2}}{2\sin\frac{x}{2}\cos\frac{x}{2}} \cdot 2\sinh x \right) }{\sqrt{3}-\sqrt{2+\cos x}}

\;=\lim\limits_{x\to 0}\;\frac{2\tan\frac{x}{2}\cdot\sinh x}{\sqrt{3}-\sqrt{2+\cos x}} \times \frac{\sqrt{3}+\sqrt{2+\cos x}}{\sqrt{3}+\sqrt{2+\cos x}}

\;=\lim\limits_{x\to 0}\;\frac{2\tan\frac{x}{2}\cdot\sinh x(\sqrt{3}+\sqrt{2+\cos x})}{1-\cos x}

\;=\lim\limits_{x\to 0}\;\frac{2\tan\frac{x}{2}\cdot\sinh x(\sqrt{3}+\sqrt{2+\cos x})}{1-\cos x}

\;=\lim\limits_{x\to 0}\;\frac{ \frac{2\sin\frac{x}{2}}{\cos\frac{x}{2}} \cdot \sinh x(\sqrt{3}+\sqrt{2+\cos x}) }{2\sin^{2}\frac{x}{2}}

\;=\lim\limits_{x\to 0}\;\frac{ \sinh x(\sqrt{3}+\sqrt{2+\cos x}) }{\cos\frac{x}{2} \cdot \sin\frac{x}{2}}

\; \because x \to 0 \Rightarrow \cos\frac{x}{2} \to 1 \;\text{ and }\; \sin\frac{x}{2} \to \frac{x}{2}

\; \sinh x \to x

\;=\;\frac{x(\sqrt{3}+\sqrt{2+1})}{\frac{x}{2}} = 2(\sqrt{3}+\sqrt{3})

=2(2\sqrt{3}) = 4\sqrt{3}

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