Limit involving integration

Question

Prove that the limit $$\lim_{x\to1}\frac{\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du}{x-1}$$ exists and equals 4.

My thoughts involving this limit are that I will have to use L'hospital's rule at some point as the denominator will equal zero if I substitute 1 in for x, but I don't know how to handle the integral in the numerator.

Answer 1

Let $f(x):=\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du$. Then

$$\lim_{x\to1}\frac{\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du}{x-1}= \lim_{x\to1}\frac{f(x)-f(1)}{x-1}=f'(1).$$

Can you proceed ?

Answer 2

Hint: $$\dfrac{d}{dx}\int_{2}^{2x}\frac{u}{1+\ln(u/2)}du=\dfrac{d}{dx}(2x) \left(\frac{2x}{1+\ln(2x/2)}\right) - \dfrac{d}{dx}(2) \left(\frac{2}{1+\ln(2/2)}\right)$$

Answer 3

Since the integrand $f$ is continuous by mean value theorem the integral in numerator is equal to $2(x-1)f(c)$ for some $c$ between $2$ and $2x$. Thus the expression under limit is equal to $2f(c)$ and as $x\to 1$ we have $c\to 2$ so that by continuity of $f$ we have the desired limit as $2f(2)=4$.