A Question on the Solution of a This Improper Integral

Question

I'm studying improper integrals, studying the equation pictured below:

When evaluating the third equation, with $\lim_{b\rightarrow\infty} -\frac{1}{u}$ with $\ln b$ as the upper limit and $u = 1$ the lower limit, wouldn't that come out to $-\frac{1}{\infty} + 1$, which I assume would come out to as a number close to 0...

I'm not sure how the third equation came out to $1$. Please explain if you can.

Thank you.

Answer 1

If $b \to \infty$, then $\log b \to \infty$. If $\log b \to \infty$, then $$\frac{1}{\log b} \to 0.$$ So you have $$\lim_{b \to \infty} \left( - \frac{1}{\log b} + 1 \right) = - 0 + 1 = 1.$$

Answer 2

$$ \lim_{b \to \infty}\left[ -\frac{1}{u} \right]_1^{\ln b} = \lim_{b \to \infty}\left[ -\left( \frac{1}{\ln b} - \frac{1}{1} \right) \right] = \lim_{b \to \infty} \left( 1 - \frac{1}{\ln b} \right) = 1 - 0 =1. $$