Sunday, 24 March 2013

real analysis - prove that a function f is continuous at x0 iff f(x0)=f(x+0)=f(x0)


prove that a function f is continuous at x0 iff lim





By definition, a function f is said to be continuous at x=x_0 if \lim\limits_{x \rightarrow x_{0}} f(x) = f(x_0)



It follows that for each \epsilon>0 there is a \delta>0 s.t. | f(x) - f(x_0)|< \epsilon whenever |x-x_0|<\delta



Considering |x-x_0|<\delta is true
-\delta < x -x_0 < \delta
x_0 - \delta < x < x_0 + \delta




Therefore,
On the right of x_0, it follows that : x_0 \leq x < x_0 + \delta such that | f(x) - f(x_0)|< \epsilon.
It shows that the function f is continuous from the right at x_0,
Therefore:
\lim\limits_{x \rightarrow x_{0}^{+}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)



Similarly, on the left of x_0, it follows that x_0 - \delta < x \leq x_0 such that | f(x) - f(x_0)|< \epsilon.
It shows that the function f is continuous from the left at x_0.
Therefore:
\lim\limits_{x \rightarrow x_{0}^{-}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)




We can conclude then that if f is continuous at x_0, f has to be continuous from both the left and the right at x_0,It follows that
\lim\limits_{x \rightarrow x_{0}^{-}} f(x) = \lim\limits_{x \rightarrow x_{0}^{+}} f(x) = \lim\limits_{x \rightarrow x_0}f(x)



Is this correct? anything could have been done differently? Any input is much appreciated

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