Skip to content
$\lim_{x\rightarrow 0}\frac{\sin x(1 – cosx)}{x^3} $
= $\lim_{x\rightarrow 0}\frac{\sin x(cos^2\frac{x}{2} + sin^2\frac{x}{2} – cos^2\frac{x}{2} + sin^2\frac{x}{2})}{x^3}$
=$ \lim_{x\rightarrow 0}\frac{\sin x\ldotp 2sin^2\frac{x}{2}}{x^3}$
= $\lim_{x\rightarrow 0}\frac{\sin x\ldotp sin^2\frac{x}{2}}{x\ldotp \frac{x^2}{4}\ldotp 2} $
= $\frac{1}{2}\lim_{x\rightarrow 0}\frac{\sin x}{x}\ldotp \lim_{x\rightarrow 0}\frac{\left(\sin \frac{x}{2}\right)^2}{\left(\frac{x}{2}\right)^2}$
= $\frac{1}{2}\lim_{x\rightarrow 0}\frac{\sin x}{x}\ldotp \left(\lim_{x\rightarrow 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2$
= $\frac{1}{2}\ldotp 1.1$
= $\frac{1}{2}$
error: Content is protected !!