Expert: Paul Klarreich - 11/17/2009

Question
Hi, I am not exactly sure how to start this prove since it seems so obvious. It is:
Prove that if f(x)= e^x then
(a) f(x) is strictly increasing for all x in R
(b) lim of x to positive infinity of f(x) = 1;
(c) lim of x to negative infinity of f(x) = 0

Answer
Questioner: Sally
Country: United States
Category: Advanced Math
Private: No
Subject: e^x

Question: Hi, I am not exactly sure how to start this prove since it seems so obvious. It is:

Prove that if f(x)= e^x then

(a) f(x) is strictly increasing for all x in R

Since  f'(x) = e^x > 0 for all x, that does it.

............................................
(b) lim of x to positive infinity of f(x) = 1;

This is more difficult.  Proving something that isn't true usually is.

HOWEVER, if you would like to prove that:

lim[x->+inf] f(x) = +inf

THAT  might be easier.  You want to show that:

Given any N > 0, no matter how large, we can find x0 sufficiently large such that  f(x) > N, whenever  x > x0.  

[That is your definition of limit.]

All you need to do is take  x0 = ln N.
Then if x > x0, e^x > e^x0 (because of part (a)) = e^(ln N) = N
................................................
(c) lim of x to negative infinity of f(x) = 0

You want to show that:

Given any epsilon > 0, no matter how SMALL, we can find -x0 sufficiently negatively large such that  f(x) < epsilon, whenever  x < -x0.  

[That is also your definition of limit.]

Just turn part (b) upside down:

Let  N = 1/epsilon.  Let x0 = ln N

Now if  x < -x0, then  x < - ln N = ln(1/N) = ln epsilon

So f(x) = e^x < e^(ln epsilon) = epsilon.

Done.