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.

https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference — evaristegd, Jul 12 '19 at 03:51
$$\lim_{b\to\infty}\dfrac{-1}u|1^{\ln b}=\lim{b\to\infty}\dfrac{-1}{\ln b}-\dfrac{-1}1=0+1$$ — J. W. Tanner, Jul 12 '19 at 03:53
as $b\to\infty, $ $\dfrac{-1}{\ln b}\to0$ so $\dfrac{-1}{\ln b}+1\to1$ — J. W. Tanner, Jul 12 '19 at 04:05

score 1 · Answer 1 · answered Jul 12 '19 at 03:52

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 $$\lim_{b \to \infty} \left( - \frac{1}{\log b} + 1 \right) = - 0 + 1 = 1.$$

answered Jul 12 '19 at 03:52

heropup

135,869

score 0 · Answer 2 · answered Jul 12 '19 at 04:56

0

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

answered Jul 12 '19 at 04:56

DDS

3,199
1
8
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A Question on the Solution of a This Improper Integral

2 Answers2