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

[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.

 

[Poopsilon]

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

 

[rock.freak667]

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

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

So as e^b tends towards infinity, you have 1/e^b 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)

 

[Robert1986]

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

Do you see it, now?

 

[SammyS]

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.

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

 

[Nano-Passion]

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

Offensive notation excused.

Thanks all for the quick reply.