# Limits Help!

• Oct 20th 2006, 07:43 AM
Limits Help!
Use the definition of limits, find the delta which corresponds to the epsilon givn the limit:
lim ( 1+x ) ^ 1/x = e ; given the epsilon = .001
x--->0

my solns:
|(1+x)^1/x - e| < .001
so,
-.001 < (1+x)^1/x - e < .001
and,
e - .001 < (1+x)^(1/x) < e + .001
.999 < (1+x)^(1/x) < 1.0010005 <=== This is wer i got stuck on how to.. ??

:(

$\displaystyle /epsilon$
• Oct 20th 2006, 10:19 AM
ThePerfectHacker
Quote:

Use the definition of limits, find the delta which corresponds to the epsilon givn the limit:
lim ( 1+x ) ^ 1/x = e ; given the epsilon = .001
x--->0

my solns:
|(1+x)^1/x - e| < .001
so,
-.001 < (1+x)^1/x - e < .001
and,
e - .001 < (1+x)^(1/x) < e + .001
.999 < (1+x)^(1/x) < 1.0010005 <=== This is wer i got stuck on how to.. ??

:(

$\displaystyle /epsilon$

We need to find a $\displaystyle \delta>0$ such that, for all
$\displaystyle |x|<\delta$ we have that,
$\displaystyle \left|\left(1+x\right)^{1/x}-e\right|<.001$
The delta inequality is more conviently written as,
$\displaystyle -\delta <x<\delta$
The second inequality can be written as,
$\displaystyle -.001<\left(1+x\right)^{1/x}-e<.001$
Thus,
$\displaystyle e-.001<\left(1+x\right)^{1/x}<e+.001$
Here is the problem, I have no idea how to do this delta-epsilon. In fact, this limit is completely proved otherwise.
Below I have attach a beautiful hand drawn diagram that shows for $\displaystyle \delta=.0001$ the delta-epsilon statement is true.

The thin red line shows the curve $\displaystyle y=(1+x)^{1/x}$
The thick blue and purple region show that it was squeezed between that area.