# Continuity of g(x) = lim{y->x}f(x)

1. Apr 18, 2013

### R_beta.v3

1. The problem statement, all variables and given/known data This problem took me a lot of time
if $g(x) = \lim_{y\rightarrow x} {f(x)}$ exist for any x, then g is continuous.
2. Relevant equations
3. The attempt at a solution
$\lim_{x\rightarrow a^+} {f(x)} = g(a)$, so if $\epsilon > 0$ then there is an $\delta_1 > 0$ such that for all x, if $a < x < \delta_1 + a$ then $|f(x) - g(a)| < \epsilon/2$. For any of those x, $lim_{y\rightarrow x^-} f(x)$ exist, so, there is a $\delta_2$ such that for all y, if $y < x < \delta_2 + y$, then $|f(y) - g(x)| < \delta/2$. Choosing some $y_0$ such that $a < y_0 < x$ we then have $|f(y_0) - g(x)| < \epsilon/2$ and $|f(y_0) - g(a)| < \epsilon/2$
So $|g(x) - g(a)| = (g(x) - f(y_0) + f(y_0) - g(a)) \leq |g(x) - f(y_0))| + |(f(y_0) - g(a)| < \epsilon/2 + \epsilon/2 = \epsilon$ Therefore, $\lim_{x\rightarrow a^+}{g(x)} = g(a)$
As x approaches a from the left its be almost the same.

Thanks

2. Apr 18, 2013

### SammyS

Staff Emeritus
It appears that you have some typos in your post ... many typos.

Do you perhaps mean to say :
If $\displaystyle \ g(y) = \lim_{x\to y} {g(x)} \$ for all y, then g is continuous.

?​

3. Apr 18, 2013

### R_beta.v3

Oh, yes, sorry, I am too used to using the x as the value approaching a, what I actually meant is the limit as {y approaches x} of f(y).

4. Apr 18, 2013

### SammyS

Staff Emeritus
You have f and g as functions, and a, x, and y as variables all over the place in your OP.

How about editing your Original Post (if it's not too late). Otherwise, type a post with the corrected version of your problem along with the solution you have so far.

What you have so far makes it virtually impossible for anyone to help you.

5. Apr 18, 2013

### R_beta.v3

Correction

That should do. Sorry for the errors, typing math symbols is not as simple nor fun as using a pen.

6. Apr 19, 2013

### SammyS

Staff Emeritus
I added some spacing, to make it a bit more readable.

I'll try to look at it in more detail soon.