Finding limit with improper integral

Question

How should I approach this question? $$\lim_{x\to0}\frac{1}{x}\int_1^{1+x}\frac{\cos t}{t} \, dt$$ I tried to use L'hospital and that gave me $-\sin(0) = 0$

The correct answer is $\cos 1$. Did I L'hospital wrong or is that the wrong way to go?

Answer 1

Assume that $$F(x) = \int_{1}^{x}\frac{\cos t}{t}\,dt.$$ By the fundamental theorem of calculus we have $F'(x)=\frac{\cos x}{x}$, and we want to compute: $$ \lim_{x\to 0}\frac{F(x+1)-F(1)}{x} \stackrel{DH}{=}\lim_{x\to 0}f(x+1) = f(1) = \cos(1).$$

Answer 2

L'Hospital's Rule works fine here. Note that we have

$$\begin{align} \lim_{x\to 0}\frac{\int_1^{1+x} \frac{\cos(t)}{t}\,dt}{x}&=\lim_{x\to 0}\frac{\frac{d}{dx}\int_1^{1+x} \frac{\cos(t)}{t}\,dt}{\frac{dx}{dx}}\\\\\ &=\lim_{x\to 0}\frac{\frac{\cos(1+x)}{1+x}}{1}\\\\ &=\cos(1) \end{align}$$

as expected!