# Limits of e functions

• Sep 29th 2008, 08:32 AM
jbecker007
Limits of e functions
Help! (Speechless)

How do I find the limits for the following:

lim x-> 0+ of e^(1/x)

Lim x->-inf of (1-e^x)/(1+e^x).

Jeannine
• Sep 29th 2008, 08:50 AM
Soroban
Hello, Jeannine!

Here's the second one . . .

Quote:

$\displaystyle \lim_{x\to\text{-}\infty}\,\frac{1-e^x}{1 + e^x}$
Divide top and bottom by $\displaystyle e^x$

$\displaystyle \lim_{x\to\text{-}\infty}\frac{\dfrac{1}{e^x} - \dfrac{e^x}{e^x}}{\dfrac{1}{e^x} + \dfrac{e^x}{e^x}} \;=\;\lim_{x\to\text{-}\infty}\frac{\dfrac{1}{e^x} - 1}{\dfrac{1}{e^x} + 1} \;=\;\frac{0-1}{0+1} \;=\;-1$

• Sep 29th 2008, 09:15 AM
jbecker007
The answer in the back of my book is not -1...
It says +1.

Jeannine
• Sep 29th 2008, 09:21 AM
Plato
You should know that $\displaystyle \color{red}\lim _{x \to - \infty } e^x = 0$

Therefore you can see that $\displaystyle \frac {1-0} {1+0} = 1$
• Sep 29th 2008, 10:19 AM
Showcase_22
what? So was Soroban wrong? I couldn't see anything wrong with that method!

Quote:

$\displaystyle \lim_{x\rightarrow 0}e^{1/x}$
Does your book say from which side x is approaching from? If x is positive, it's $\displaystyle +\infty$ and if x is negative it's $\displaystyle -\infty$.

Someone good at analysis might find a way around this though.
• Sep 29th 2008, 10:25 AM
Jhevon
Quote:

Originally Posted by Showcase_22
what? So was Soroban wrong? I couldn't see anything wrong with that method!

Does your book say from which side x is approaching from? If x is positive, it's $\displaystyle +\infty$ and if x is negative it's $\displaystyle -\infty$.

Someone good at analysis might find a way around this though.

$\displaystyle \lim_{x \to - \infty} \frac 1{e^x} = \lim_{x \to - \infty} e^{-x} = \infty$ not zero. i think Soroban forgot about the negative infinity.

one-sided limits don't exist if the limit is infinite. how could you approach infinity from the right? or - infinity from the left?
• Sep 29th 2008, 10:37 AM
Showcase_22
Quote:

one-sided limits don't exist if the limit is infinite. how could you approach infinity from the right? or - infinity from the left?
I meant x was approaching zero from the right or left. I should have explained it better.

$\displaystyle \lim_{x \rightarrow 0+} \frac {1}{x}=+ \infty$

$\displaystyle \lim_{x \rightarrow 0-} \frac {1}{x}=- \infty$

(I think I wrote that out correctly :s)

Then this would make:

$\displaystyle e^{(+ \infty)}= \infty$

$\displaystyle e^{(- \infty)}=0$

Oops! Wrong in my previous post!

Am I making the problem too difficult?

Quote:

$\displaystyle \lim_{x \rightarrow -\infty} e^{(-x)}$
I didn't notice it was $\displaystyle - \infty$. I thought it was $\displaystyle + \infty$. Oops! (Lipssealed)
• Sep 29th 2008, 10:46 AM
Jhevon
Quote:

Originally Posted by Showcase_22
I didn't notice it was $\displaystyle - \infty$. I thought it was $\displaystyle + \infty$. Oops! (Lipssealed)

yes. i suppose Soroban forgot that fact as well, as i said in my last post :D
• Sep 29th 2008, 10:58 AM
bkarpuz
You may also do substitution in limit as follows:
$\displaystyle \lim\limits_{x\to-\infty}\mathrm{e}^{x}\stackrel{y\mapsto-x}{=}\lim\limits_{y\to\infty}\mathrm{e}^{-y}=\lim\limits_{y\to\infty}\frac{1}{\mathrm{e}^{y} }=0$.
• Sep 29th 2008, 11:20 AM
Plato
Look all of you; the discussion was about the second of the two problems not the fiist: $\displaystyle \lim _{x \to - \infty } \frac{{1 - e^x }}{{1 + e^x }}$.
That limit can easily be done by simple inspection knowing that $\displaystyle \lim _{x \to - \infty } e^x = 0$.

The first problem was $\displaystyle \lim _{x \to 0^ + } e^{\frac{1}{x}}$ which is also easy to see that the limit does exist or is $\displaystyle +\infty$ depending on the textbook.