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

  • Thread starter R_beta.v3
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  • #1
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Homework Statement

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

Homework Equations


The Attempt at a Solution


[itex]
\lim_{x\rightarrow a^+} {f(x)} = g(a)
[/itex], 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
 

Answers and Replies

  • #2
SammyS
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Homework Statement

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

Homework Equations


The Attempt at a Solution


[itex]
\lim_{x\rightarrow a^+} {f(x)} = g(a)
[/itex], 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.

?​
 
  • #3
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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
SammyS
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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.
 
  • #5
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Correction

Homework Statement

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

Homework Equations


The Attempt at a Solution


[itex]
\lim_{x\rightarrow a^+} {f(x)} = g(a)
[/itex], 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.
 
  • #6
SammyS
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Science Advisor
Homework Helper
Gold Member
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Homework Statement

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

Homework Equations


The Attempt at a Solution


[itex]\displaystyle
\lim_{x\rightarrow a^+} {f(x)} = g(a)
[/itex], 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.
 

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