🛠️ JEE➗ Maths

The value of limit: \(\lim _{x\to 0}(\csc x)\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right)\text{…

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The value of limit: \(\lim _{x\to 0}(\csc x)\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right)\text{ is }\)

a

0

b

1

c

\(\frac{1}{2\sqrt{5}}\)

d

\(\frac{-1}{2\sqrt{5}}\)

✓ Correct answer: d)

\(\frac{-1}{2\sqrt{5}}\)

Explanation

\(\lim _{x\to 0}\csc x\left(\sqrt{2{\cos }^{2}x+3\cos x}-\sqrt{{\cos }^{2}x+\sin x+4}\right)\\ =\lim _{x\to 0}\frac{1}{\sin x}\left[\frac{2{\cos }^{2}x+3\cos x-{\cos }^{2}x-\sin x-4}{\sqrt{2{\cos }^{2}x+3\cos x}+\sqrt{{\cos }^{2}x+\sin x+4}}\right]\\ =\lim _{x\to 0}\frac{1}{2\sqrt{5}}(\frac{{\cos }^{2}x+3\cos x-\sin x-4}{\sin x})\\ \mathrm{using}\mathrm{L}'\mathrm{hospital}\mathrm{rule}-\\ =\frac{1}{2\sqrt{5}}\lim _{x\to 0}\left(-\frac{2\cos x\sin x-3\sin x-\cos x}{\cos x}\right)\\ =\frac{-1}{2\sqrt{5}}\\ \\\)

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