It looks like it should be simple, but I'm having some trouble wrapping my head around this one. Any help would be greatly appreciated.
$\displaystyle \lim_{x\rightarrow3}\bigg(\frac{x}{x-3}\int_{3}^{x}\frac{\sin{t}}{t} dt\bigg)$
Thanks.
Using the usual algebra of limits you have $\displaystyle \left( \lim_{x\rightarrow3} x\right) \cdot \lim_{x \rightarrow 3} \frac{\int_{3}^{x}\frac{\sin{t}}{t} dt}{x - 3}$.
The second limit has the indeterminant form 0/0 so apply l'Hopitals to it (use the Fundamental Theorem of Calculus to differentiate the numerator).
Let F(t) be an antiderivative of $\displaystyle \frac{\sin t}{t} $
then $\displaystyle \lim_{x \to 3} \frac{x}{x-3} \int^{x}_{3} \frac{\sin t}{t} \ dt = \lim_{x \to 3} \frac{x}{x-3} \big(F(x)-F(3)\big) $
$\displaystyle = \lim_{x \to 3} x \cdot \lim_{x \to 3} \frac{F(x)-F(3)}{x-3} $
$\displaystyle = 3F'(3) = 3 \ \frac{\sin 3}{3} = \sin 3 $