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Question

Mathematics Question on limits and derivatives

lim⁡x→0sin⁡2x2−1+cos⁡x\displaystyle\lim_{x\to0} \frac{\sin^{2}x}{\sqrt{2} - \sqrt{1+\cos x}} equals :

A

222 \sqrt{2}

B

424 \sqrt{2}

C

2 \sqrt{2}

D

4

Answer

424 \sqrt{2}

Explanation

Solution

lim⁡x→0(sin⁡2xx2)(2+1+cos⁡x)(1−cos⁡xx2)\lim_{x\to0} \frac{\left(\frac{\sin^{2}x}{x^{2}}\right)\left(\sqrt{2} + \sqrt{1+\cos x}\right)}{\left(\frac{1-\cos x}{x^{2}}\right)}
=(1)2.(22)12=42= \frac{\left(1\right)^{2} .\left(2\sqrt{2}\right)}{\frac{1}{2}} = 4\sqrt{2}