# Prove that lim_{x to x_0} f(x) = L if and only if lim_{x to 0} f(x+x_0) = L. Assume

• Feb 28th 2010, 06:42 PM
tn11631
Prove that lim_{x to x_0} f(x) = L if and only if lim_{x to 0} f(x+x_0) = L. Assume
second to last question.

Prove that lim_{x to x_0} f(x) = L if and only if lim_{x to 0} f(x+x_0) = L. Assume that L is finite.
• Mar 1st 2010, 03:17 AM
HallsofIvy
Quote:

Originally Posted by tn11631
second to last question.

Prove that lim_{x to x_0} f(x) = L if and only if lim_{x to 0} f(x+x_0) = L. Assume that L is finite.

That's pretty straight forward. In terms of the definitions of those limits you want to prove that
"Given $\epsilon> 0$ there exist $\delta> 0$ such that if $|x-x_0|< \delta$ then $|f(x)- L|< \epsilon$"

if and only if
"Given $\epsilon> 0$ there exist $\delta> 0$ such that if $|y|< \delta$ then $|f(x_0+ y)- L|< \epsilon$.
(I've changed "x" to "y" in the second part so as not to confuse the two uses of "x".)

Comparing those two, they will be the same if $x_0+ y= x$. In other words, let $y= x- x_0$.
• Mar 1st 2010, 08:53 AM
tn11631
In other words, let y= x- x_0.....do i have to show anything more or do the two definitions take of it? Sorry i'm really bad at proofs..
• Mar 1st 2010, 12:46 PM
Drexel28
Quote:

Originally Posted by tn11631
In other words, let y= x- x_0.....do i have to show anything more or do the two definitions take of it? Sorry i'm really bad at proofs..

Do exactly as HallsOfIvy said. The basic idea is that in the limit $\lim_{x\to x-0}f(x)=0$ if we let $z=x-x_0$ then we get $\lim_{z\to 0}f(z)=0$. Formalize this.