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

**tn11631** · February 28th 2010, 06:42 PM

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.

**HallsofIvy** · March 1st 2010, 03:17 AM

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 $y= x- x_0$ .

**tn11631** · March 1st 2010, 08:53 AM

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..

**Drexel28** · March 1st 2010, 12:46 PM

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.

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