l i m x → 0 ∫ 0 x t s i n 10 t d t x , is equal to

We have, $I=\underset{x\to 0}{\lim }\frac{{\int }_0^{x}t\sin \left(10t\right)dt}{x}$

The given limiting form is the indeterminate form $\frac{0}{0}$.

Thus, using Newton-Leibnitz Rule (differentiation under integral sign) and L' Hospital Rule, we get, 

$I=\underset{x\to 0}{\lim }\frac{{\displaystyle \frac{d}{dx}}\left\{{\int }_0^{x}t\sin \left(10t\right)dt\right\}}{{\displaystyle \frac{d}{dx}}\left(x\right)}$

$I=\underset{x\to 0}{\mathrm{l}\mathrm{i}\mathrm{m}}\frac{x\sin \left(10x\right)}{1}$

$=0$.