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

## 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.

## 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

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SammyS
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## 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.

## 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.

?​

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
Staff Emeritus
Homework Helper
Gold Member
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.

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.

## 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
Staff Emeritus
Homework Helper
Gold Member

## 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.

## 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.