Question

Evaluate the limit:

$\underset{x\to 0}{lim}\frac{\sqrt{2}-\sqrt{1+\cos x}}{si{n}^{2}x}$

Answer 1

We have,

$\underset{x\to 0}{lim}\frac{\sqrt{2}-\sqrt{1+\cos x}}{si{n}^{2}x}$

It becomes 0/0 form.

On rationalising the numerator, we get

$=\underset{x\to 0}{lim}\frac{\sqrt{2}-\sqrt{1+cosx}}{si{n}^{2}x}\times \frac{\sqrt{2}+\sqrt{1+\cos x}}{\sqrt{2}+\sqrt{1+\cos x}}$

$=\underset{x\to 0}{lim}\frac{(\sqrt{2}{)}^2-(\sqrt{1+\cos x}{)}^2}{si{n}^{2}x(\sqrt{2}+\sqrt{1+\cos x})}$

$[\because (a+b\left)\right(a-b)=a^2-b^2]$

$=\underset{x\to 0}{lim}\frac{2-1-\cos x}{{\sin }^{2}x(\sqrt{2}+\sqrt{1+\cos x})}$

$[\because 1+\cos x=2{\cos }^2\frac{x}{2}]$

$=\underset{x\to 0}{lim}\frac{1-\cos x}{{\sin }^{2}x(\sqrt{2}+\sqrt{2{\cos }^2\frac{x}{2}})}$

$=\underset{x\to 0}{lim}\frac{2{\sin }^2\frac{x}{2}}{\sqrt{2}si{n}^{2}x(1+cos\frac{x}{2})}$

$[\because 1-\cos x=2{\sin }^2\frac{x}{2}]$

$=\underset{x\to 0}{lim}\frac{\frac{2{\sin }^2\frac{x}{2}}{\frac{x^2}{4}}\times \frac{x^2}{4}}{\frac{\sqrt{2}si{n}^{2}x}{x^2}\times x^2(1+\cos \frac{x}{2})}$

$=\frac{1}{2\sqrt{2}(1+1)}$

$=\frac{1}{4\sqrt{2}}$