limit of integrals

$\begin{aligned} L & = \lim_{x \to 3} \frac {x\int_3^x \frac {\sin t}t dt}{x-3} & \small \blue{\text{A 0/0 case, L'Hôpital's rule applies.}} \\ & = \lim_{x \to 3} \frac {\int_3^x \frac {\sin t}t dt+x \times \frac {\sin x}x}{1} & \small \blue{\text{Differentiate up and down w.r.t. }x} \\ & = \lim_{x \to 3} \sin x = \sin 3 \end{aligned}$

Therefore, $A=\boxed 3$ .

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Reference: L'Hôpital's rule

My solution see the hole limit as a derivative

Let $F(x)=\displaystyle\int_a^x \dfrac{\sin(t)}{t} dt$

By fundamental theorem of analysis, $F'(x)= \dfrac{\sin(x)}{x}$

So,

$\begin{aligned} \underset{x\to 3}{\lim} \ \dfrac{1}{x-3}\displaystyle\int_3^x \dfrac{\sin(t)}{t} dt & = \underset{x\to 3}{\lim} \ \dfrac{F(x)-F(3)}{x-3} \\ & = F'(3)\\ & = \frac{\sin(3)}{3} \end{aligned}$

Thus,

$\begin{aligned} \underset{x\to 3}{\lim} \ x \times \dfrac{1}{x-3}\displaystyle\int_3^x \dfrac{\sin(t)}{t} dt & = 3 \times \underset{x\to 3}{\lim} \ \dfrac{1}{x-3}\displaystyle\int_3^x \dfrac{\sin(t)}{t} dt \\ & = 3 \times \frac{\sin(3)}{3}\\ & = \boxed{\sin(3)} \end{aligned}$