The value of the following limit without using L'Hopital

Find the limit
$$
\lim_{x\to 3}\left(\frac{x}{x-3}\int_{3}^{x}\frac{\sin t}{t} dt\right)
$$
without using L'Hopital's rule.

Answer

By the Mean Value Theorem,
$$\int_3^x\frac{\sin t}{t}=(x-3) \frac{\sin c_x}{c_x}$$ for some $c_x$ between $3$ and $x$. So our product is equal to
$$x\cdot\frac{\sin c_x}{c_x}.$$
As $x\to 3$, $\sin{c_x}\to\sin 3$ and $c_x\to 3$, so our limit is $\sin 3$.