Prove limit of integral

Question

Prove that

$$\lim_{x\to3}\frac{x}{x-3}\int^x_3\frac{\sin(t)}{t}\,\mathrm{d}t=\sin(3).$$

I've tried to split the limit in two, but that doesn't help since I can't find the antiderivative of the integral. Any pointers would be greatly appreciated!

Answer 1

\begin{align*} \lim_{x\rightarrow 3}\dfrac{x}{x-3}\int_{3}^{x}\dfrac{\sin t}{t}dt&=\lim_{x\rightarrow 3}x\cdot\dfrac{\sin\eta_{x}}{\eta_{x}},~~~~\eta_{x}~\text{in between $3$ and $x$}\\ &=3\cdot\dfrac{\sin 3}{3}\\ &=\sin 3, \end{align*} where we have used the Mean Value Theorem for integrals.

Answer 2

Simply use Hopital theorem.It is a $0/0$ limit.

Answer 3

Well, by fundamental theorem of calculus we have $$\lim_{x\to 3}\frac{1}{x-3}\int_{3}^{x}\frac{\sin t} {t} \, dt=\frac{\sin 3}{3}$$ and hence the desired limit is equal to $3\cdot\dfrac{\sin 3}{3}=\sin 3$.