Question

Solve without L hospital​

Answer 1

Step-by-step explanation:

hope this helps ...

i said i could do this whole day...it doesn't mean...

I have corrected it

Answer 2

Answer:

$\frac{\sqrt{2}}{8}$

Step-by-step explanation:

$\lim_{x\rightarrow0}\frac{\sqrt2-\sqrt{1+\cos x}}{\sin^2x}\\=\lim_{x\rightarrow0}\bigg(\frac{\sqrt2-\sqrt{1+\cos x}}{\sin^2x}\cdot\frac{\sqrt2+\sqrt{1+\cos x}}{\sqrt2+\sqrt{1+\cos x}}\bigg)\\=\lim_{x\rightarrow0}\bigg(\frac{1-\cos x}{\sin^2x}\cdot\frac{1}{\sqrt2+\sqrt{1+\cos x}}\bigg)\\= \lim_{x\rightarrow0}\frac{1-\cos x}{1-\cos^2 x}\cdot\lim_{x\rightarrow0}\frac{1}{\sqrt2+\sqrt{1+\cos x}}\\=\lim_{x\rightarrow0}\frac{1}{1+\cos x}\cdot\lim_{x\rightarrow0}\frac{1}{\sqrt2+\sqrt{1+\cos x}}$

$=\frac12\cdot\frac{1}{\sqrt2+\sqrt2}\\=\frac{\sqrt2}{8}$