AIEEE 2002 (Offline)Question 1
$\lim_{x\to0}\frac{\sqrt{1-\cos2x}}{\sqrt2\,x}$ is
Limits & Continuity
Solution
$\frac{\sqrt{1-\cos2x}}{\sqrt2x}=\frac{|\sin x|}{x}$, which tends to $1$ from the right and $-1$ from the left; the limit does not exist.Solution
$\frac{\sqrt{1-\cos2x}}{\sqrt2x}=\frac{|\sin x|}{x}$, which tends to $1$ from the right and $-1$ from the left; the limit does not exist.