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Limits Continuity and Differentiability

Section: Mathematics

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Question : 28 of 61

Marks: +1, -0

\lim\limits_{x\to 0}\;\frac{(\sqrt{2})-\sqrt{1+\cos x}}{\sqrt{15+\cos 2x}-4}=

[AP EAPCET 21 May 2025 S1]

$-\;\frac{1}{\sqrt{2}}$
$\;\frac{1}{\sqrt{2}}$
$\sqrt{2}$
$-\sqrt{2}$

Solution:

\lim\limits_{x\to \infty}\;\frac{\sqrt{2}-\sqrt{1+\cos x}}{\sqrt{15+\cos 2x-4}}

By rationalizing

N^{r}

and

D^{r}

, we get

\lim\limits_{x\to 0}\;\frac{2-1-\cos x}{\sqrt{2}+\sqrt{1+\cos x}}\times\frac{\sqrt{15+\cos 2x}+4}{15+\cos 2x-16}

\;=\lim\limits_{x\to 0}\;\frac{1-\cos x}{-(1-\cos 2x)}\times\;\frac{4+4}{\sqrt{2}+\sqrt{2}}

\;=-\;\frac{8}{2\sqrt{2}}\lim\limits_{x\to 0}\;\frac{\frac{1-\cos x}{x^{2}}\times x^{2}}{1-\frac{\cos 2x}{(2x)^{2}}\times 4x^{2}}

Using result

\lim\limits_{x\to 0}\;\frac{1-\cos x}{x^{2}}=\;\frac{1}{2}

=-\;\frac{8}{2\sqrt{2}}\times\;\frac{\frac{1}{2}}{\frac{1}{2}}\times\;\frac{1}{4}=-\;\frac{1}{\sqrt{2}}

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