Limit with two factors, one rational and one integral. $$\lim_{x\rightarrow\pi}\frac{x}{x-\pi}\int^x_\pi\frac{\cos t}{t}$$
I'm trying to figure out how to deal with this one. I know that $\int^a_af = 0$. However, the problem comes from the factor with its denominator going to $0$ when $x\rightarrow\pi$.
What can I do to fix this one?
 A: Since 
$$\lim_{x\to\pi}\int_\pi^x\frac{\cos t}{t}=0\mbox{ and }\lim_{x\to\pi}(x-\pi)=0,$$by L'Hospital's rule, we have 
$$\lim_{x\to\pi}\frac{\displaystyle\int_\pi^x\frac{\cos t}{t}}{x-\pi}=\lim_{x\to\pi}\frac{\displaystyle(\frac{\cos x}{x})}{1}=\frac{\cos\pi}{\pi}=-\frac{1}{\pi}.$$
Therefore, we have 
$$\lim_{x\rightarrow\pi}\frac{x}{x-\pi}\int^x_\pi\frac{\cos t}{t}
=\left(\lim_{x\rightarrow\pi}x\right)\left(\lim_{x\to\pi}\int_\pi^x\frac{\cos t}{t}\right)
=\pi\cdot(-\frac{1}{\pi})=-1.$$
A: Set $$f(x) := x \int_{\pi}^x \frac{\cos t}{t} dt.$$ Then, by definition,
$$\lim_{x \to \pi} \frac{x}{x - \pi} \int_{\pi}^x \frac{\cos t}{t} dt = \lim_{x \to \pi} \frac{f(x) - f(\pi)}{x - \pi} = f'(\pi).$$
This derivative can be evaluated directly with the product rule and the Fundamental Theorem of Calculus.
A: This can be done also with mean value theorem for integrals: .
$$ \int_{\pi}^x \frac{\cos t}{t} dt = \frac{\cos \xi}{\xi} (x-\pi) \text{ , where } \xi \in(\pi,x). $$
When this is placed into the expression, we get:
$$\lim_{x \rightarrow \pi} \frac{x}{x-\pi} \frac{\cos \xi}{\xi} (x-\pi) = \lim_{x \rightarrow \pi}x\frac{\cos \xi}{\xi} ~ \text{ , where } \xi \in(\pi,x)$$
and now, when $x$ approaches the $\pi$, $\xi$ is squeezed to $\pi$ and we get the same result as when L'Hôpital's rule is applied: $-1$.
