# Thread: function proof

1. ## function proof

(2) Show that lim_{x \to x_0} f(x) = L if and only if lim_{x \to 0} f(x+x_0) = L. Assume x_0 and L are finite.
(3) Show that if lim_{x \to x_0} f(x) = L and E is a set which has x_0 as an accumulation point, then lim_{x \to x_0, x in E} f(x) = L. Give an example to show that the converse may fail. Assume x_0 and L are finite.

can anyone solve any of these?

2. You know that if you could post using even basic LaTeX, we could understand what you are asking. It is very hard to know what you are asking.

3. what do u mean basic latex?

4. Here is what you asked in basic LaTex. Incidentally, what you typed was very close to LaTex code, add a few "\"'s and put math tags around it and you would have:

Originally Posted by ruprotein
(2) Show that $\lim_{x \to x_0} f(x) = L$ if and only if $\lim_{x \to 0} f(x+x_0) = L$. Assume $x_0$ and $L$ are finite.

(3) Show that if $\lim_{x \to x_0} f(x) = L$ and $E$ is a set which has $x_0$ as an accumulation point, then $\lim_{x \to x_0} f(x) = L$, $x \in E$. Give an example to show that the converse may fail. Assume $x_0$ and $L$ are finite.

can anyone solve any of these?

5. Originally Posted by ruprotein
(2) Show that lim_{x \to x_0} f(x) = L if and only if lim_{x \to 0} f(x+x_0) = L. Assume x_0 and L are finite.
$\lim_{x\to x_0} f(x) = L$ means:
$0<|y-x_0| < \delta \implies |f(x) - L| < \epsilon$.
Now if $y = x_0+x$ where $0<|x|<\delta$ we have:
$|y-x_0| = |x_0+x - x_0| = |x| \in \left( 0, \delta \right)$
Thus,
$|f(x+x_0) - L| < \epsilon$.

6. thanks a lot has anyone ghot the other one?

Show if lim f(x) = L then if E is any set which has x_o as an accumulation point , lim f(x) = L. But show converse fails.

7. Originally Posted by ruprotein
thanks a lot has anyone ghot the other one?

Show if lim f(x) = L then if E is any set which has x_o as an accumulation point , lim f(x) = L. But show converse fails.
Is $E = \{ f(x) \}$?

8. the guy above restated the question in Latex form