The limit of $ m\int_{a}^{1/m} \frac{dx}{x}=0 $ and $ m\int_{a}^{\infty} \frac{dx}{x^{1+m}}=0$ as $m\to0$

Question

Given $ a >0 $ is it correct that $$ \lim_{m\to 0}m\int_{a}^{1/m} \frac{dx}{x}=0 $$ by the properties of the logarithm function?

Or on the other hand, $$\lim_{m\to 0} m\int_{a}^{\infty} \frac{dx}{x^{1+m}}=0$$

Do both of these hold because $ x\log(x)\to 0 $ as $x \to 0 $?

Answer 1

We need to be careful. Neither integral exists if $m\lt 0$. But it does make sense to compute the limits as $m$ approaches $0$ from the right.

We look for example at $$\lim_{m\to 0^+} m\int_a^\infty \frac{dx}{x^{1+m}}.$$ Calculate. The integral is equal to $\frac{1}{m} a^{-m}$. So we want $$\lim_{m\to 0^+} a^{-m}.$$ This limit is $1$.

For the first problem, again, for positive $m$ we integrate. We want to calculate $$\lim_{m\to 0^+} m(\ln(1/m)-\ln(a)).$$ Equivalently, we want $$\lim_{m\to 0^+}(-m)(\ln m +\ln a).$$ By the result you quoted, the limit is $0$.