# How do you prove that the exponential function is continuous at x=0 and how do you prove it is continuous for all real x?

How do you prove that the exponential function is continuous at x=0 and how do you prove it is continuous for all real x?
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faux0101d
Suppose that exp x is continuous at x=0. This means that
$\underset{h\to 0}{lim}{e}^{h}=1$
Now, consider any other x. Then
${e}^{x}\cdot \underset{h\to 0}{lim}{e}^{h}={e}^{x}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{lim}{e}^{x}\cdot {e}^{h}={e}^{x}\phantom{\rule{0ex}{0ex}}\underset{h\to 0}{lim}{e}^{x+h}={e}^{x}$
which says this is continuous at every other point. The same idea works in showing it is differentiable at every point as soon as it is differentiable at the origin. Now, it depends on how you define it to devise a proof of continuity and/or differentiability at the origin.
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Cory Russell
If you know that the exponential function is differntiable on R,
$\left(\mathrm{exp}{\right)}^{\prime }\left(x\right)=\mathrm{exp}\left(x\right),\phantom{\rule{1em}{0ex}}\mathrm{\forall }x\in \mathbb{R},$
then it's continuous on R as well.