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Take the function $f(x) = e^{x}$ and I want to prove this is continuous at $x=0$ using epsilon-delta definition.

ie. $\forall \epsilon>0, \exists\delta>0 \text{ s.t. } |x - 0| < \delta \implies |e^{x}-1| < \epsilon$

So $|e^{x}-1| < |e^{x}| + |1| = e^{x}+1$

Now $|x - 0| < \delta$ or $|x|<\delta$ so $x<\delta$

Hence $e^{x}+1 < e^{\delta}+1$

Take $\epsilon = e^{\delta}+1$ hence $\delta = \ln{(\epsilon -1)}$. But this doesn't show it because if we choose an $\epsilon \le 2$ we can't find a $\delta > 0$ such that the implication holds.

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    $\begingroup$ Right: Although $|e^x - 1| \le e^x + 1$ is completely true, it's not useful here. This really depends on how you've defined $e^x$, since the proof it's continuous uses different methods depending on the definition. $\endgroup$ – user61527 Apr 30 '14 at 22:45
  • $\begingroup$ Using the inequality $|e^x-1|<e^x+1$ is a bad way to try to prove, cause this is always bigger then $1$, and you need something arbitrarily small. $\endgroup$ – Integral Apr 30 '14 at 22:46
  • $\begingroup$ So how would I tackle this then? (using epsilon delta not comparison) $\endgroup$ – user2850514 Apr 30 '14 at 22:47
  • $\begingroup$ @user2850514 As T. said, you need to tells us what's your definition of $e^x$ (there are several ways to define it). From this definition, the trick is to relate $|e^x-1|$ with $|x|$ (you didn't do this). $\endgroup$ – Git Gud Apr 30 '14 at 22:50
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    $\begingroup$ @user2850514 Reread my comment, I said there are several ways to define $e^x$. It's not that $e^x$ has several meanings, it's that there are several ways to define the same thing. $\endgroup$ – Git Gud Apr 30 '14 at 22:54
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You could use the following $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots$$

Therefore,

$$|e^x -1| = \left|x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots\right|$$

Taking $|x-0| < \delta$, where $0<\delta<1$ and $\frac{\delta}{1-\delta}<\epsilon$, we have that $$|e^x -1| = \left|x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \right| < \delta + \frac{\delta^2}{2!} + \frac{\delta^3}{3!} + \ldots<$$ $$< \delta + \delta^2 + \delta^3 + \ldots = \frac{\delta}{1-\delta} < \epsilon.$$

PS: if you don't know this definition for $e^x$, disregard my answer.

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  • $\begingroup$ I want to do it for the function $e^{x}$ without substituting the polynomial. I have already practised polynomials and now I want to try exponentials. $\endgroup$ – user2850514 Apr 30 '14 at 23:05
  • $\begingroup$ Cool, but since an infinite sum is actually a limit the strict inequality needs to be justified. This a consequence of $$\left|x + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \right|\leq |x|+\left|\frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \right|<\delta +\delta ^2\ldots$$ $\endgroup$ – Git Gud Apr 30 '14 at 23:07
  • $\begingroup$ You are right, but looks like my answer is not what he (or she) wants. $\endgroup$ – Integral Apr 30 '14 at 23:08
  • $\begingroup$ @Integral He doesn't know what he wants. If you read the comments in the question you'll see this. I'm trying to explain. $\endgroup$ – Git Gud Apr 30 '14 at 23:09
  • $\begingroup$ Well...that's a problem! Im gonna check, and thanks for the observation. $\endgroup$ – Integral Apr 30 '14 at 23:10

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