# 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?
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it

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.
###### Did you like this example?
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.