# 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?

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: $-1$.

• In the second equality, where did the factor $\xi$ come from? Nvm, saw you just missed a comma :) – PurpleManiac Nov 30 '14 at 12:28

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 $$\lim_{x\rightarrow\pi}\frac{x}{x-\pi}\int^x_\pi\frac{\cos t}{t} =\left(\lim_{x\rightarrow\pi}x\right)\left(\lim_{x\to\pi}\int_\pi^x\frac{\cos t}{t}\right) =\pi\cdot(-\frac{1}{\pi})=-1.$$

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.