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.