# Lim as b approaches infinite -e^(-b) = 0 ? 0__o

1. Apr 1, 2012

### Nano-Passion

I'm doing improper integrals and my book is saying that

lim as b approaches infinite of
$$-e^{-b} + 1$$
$$= 1$$

This doesn't make any sense to me. I would think that it would be

$$-e^{-b} + 1$$
$$= - ∞ + 1$$
$$= - ∞$$

Help is appreciated.

2. Apr 1, 2012

### Poopsilon

$$-e^{-b} = \frac{-1}{e^b}$$Thus you have a fraction whose denominator is getting really large.

3. Apr 1, 2012

### rock.freak667

As b→∞, eb→∞ as well. That part you know.

The part that you are missing is that e-b=1/eb.

So as eb tends towards infinity, you have 1/eb tending towards zero.

(If you need to check, calculate 1/e,1/e2,1/e5.1/e100 and you will see it gets smaller and smaller until it approaches zero)

4. Apr 1, 2012

### Robert1986

$b$ is getting really big, right? Doesn't $e^{-b} = \frac{1}{e^b}$.

Do you see it, now?

5. Apr 1, 2012

### SammyS

Staff Emeritus
$\displaystyle e^{-b}= \frac{1}{e^{b}}\ \ \to\ \ \frac{1}{\infty}\,.$ (Please excuse the offensive notation.)

6. Apr 1, 2012

### Nano-Passion

Offensive notation excused.

Thanks all for the quick reply.