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

[R_beta.v3]

Homework Statement

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.

Homework Equations

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

 

[SammyS]

Homework Statement

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.

Homework Equations

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

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.

?​

 

[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).

 

[SammyS]

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).

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.

 

[R_beta.v3]

Correction

Homework Statement

This problem took me a lot of time
if $g(x) = \lim_{y\rightarrow x} {f(y)}$ exist for any x, then g is continuous.

Homework Equations

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, $lim_{y\rightarrow x^-} f(y)$ exist, so, there is a $\delta_2 > 0$ 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 almost the same.

Thanks

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

 

[SammyS]

Homework Statement

This problem took me a lot of time
if $\displaystyle g(x) = \lim_{y\,\to\, x} {f(y)}$ exist for any x, then g is continuous.

Homework Equations

The Attempt at a Solution

$\displaystyle \lim_{x\rightarrow a^+} {f(x)} = g(a)$, so if $\displaystyle \epsilon > 0$ then there is an $\delta_1 > 0$ such that for all x, if $\displaystyle a < x < \delta_1 + a$ then $|f(x) - g(a)| < \epsilon/2$.

For any of those x, $\displaystyle \lim_{y\rightarrow x^-} f(x)$ exist, so, there is a $\displaystyle \delta_2 >0$ 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 $\displaystyle |f(y_0) - g(x)| < \epsilon/2$ and $|f(y_0) - g(a)| < \epsilon/2\ .$

So $\displaystyle |g(x) - g(a)| = (g(x) - f(y_0) + f(y_0) - g(a)) \leq |g(x) - f(y_0))| + |(f(y_0) - g(a)| < \epsilon/2 + \epsilon/2 = \epsilon\ .$

Therefore, $\displaystyle \lim_{x\rightarrow a^+}{g(x)} = g(a)\ .\$

As x approaches a from the left it's almost the same.

Thanks

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

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

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