Question

$\underset{x\to 0}{lim}\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}$

Answer 1

$\underset{x\to 0}{lim}\frac{(1-\cos 2x)\sin 5x}{x^2\sin 3x}$

we know that

$co{s}^{2}x=1-2{\sin }^{2}x$ $2{\sin }^{2}x=1-\cos x$

$\underset{x\to 0}{lim}\frac{2{\sin }^{2}x\sin 5x}{x^2\sin 3x}$

we know that $\frac{\sin x}{x}=1$ at $x\to 0$

$\underset{x\to 0}{lim}2{\left(\frac{\sin x}{x}\right)}^2\times \left(\frac{\sin 5x}{5x}\right)\times \left(\frac{3x}{\sin 3x}\right)\times \frac{5x}{3x}$

$\underset{x\to 0}{lim}2\times \frac{5}{3}=\frac{10}{3}$