Limit problem by Hosam Hajjir

Multiply numerator and denominator by $1 + \cos(x)$ . The fraction in question then reduces to $\dfrac{\sin(x)}{x} \times \dfrac{1}{1+ \cos(x)}$ , which goes to $\boxed{0.5}$ as $x \to 0$ .

$\displaystyle \lim_{x \to 0} \frac{1-\cos(x)}{x\sin(x)}$

If we put $x = 0$ into the limit, we will get indeterminate form of type $\frac{0}{0}$ . Apply L'Hôpital's rule :

$\displaystyle \lim_{x \to 0} \frac{1-\cos(x)}{x\sin(x)} = \lim_{x \to 0} \frac{\frac{d}{dx} (1-\cos(x)) }{ \frac{d}{dx} (x\sin(x)) }$

$\displaystyle = \lim_{x \to 0} \frac{\sin(x)}{x\cos(x)+\sin(x)}$

$\displaystyle = \lim_{x \to 0} \frac{\frac{\sin(x)}{x}} {\cos(x)+ \frac {\sin(x)}{x}}$

$= \frac{\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x}}{\displaystyle \lim_{x \to 0} \cos(x) + \lim_{x \to 0} \frac{\sin(x)}{x}}$

We know that $\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x} = 1$

$\implies \frac{\displaystyle \lim_{x \to 0} \frac{\sin(x)}{x}}{\displaystyle \lim_{x \to 0} \cos(x) + \displaystyle \lim_{x \to 0} \frac{\sin(x)}{x}} = \displaystyle \frac{1}{\cos(0) + 1} = \boxed{\frac{1}{2}}$

@Brian Charlesworth has a very elegant answer, but just to show an alternative approach:

For small $x$ , we have $\sin x \approx x$ and $\cos x \approx 1-\frac12 x^2$ . Substituting these in we get the limit $\frac{\frac12 x^2}{x^2}=\boxed{\frac12}$

\lim {x \to 0}\frac{1-cosx}{xsinx}=\lim {x \to 0}\frac{2sin^2(x/2)}{x^2}=\lim_{x \to 0}2\frac{x^2}{4x^2}=\frac{1}{2}

Assuming sinx reaches to x (As x reaches to 0) And sin^2\frac{x}{2 } reaches to (\frac{x}{2})^2