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## Homework Statement

This problem took me a lot of timeif [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