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The function $f$ is given by

$$ f(x) = \begin{cases} e^{−\frac{1}{|x|}} & \text{if } x \ne 0\\ p & \text{if } x = 0 \end{cases}. $$

I have to find out which value of $p$ I need, so that the function is differentiable.

Left limit:

so what is the limit $$\lim_{x\to 0^{-}} -\frac{e^{-\frac{1}{x}}}{x^2}.$$

i know $e^{-1/x}$ limit is $0$ so is the total limit then also $0$?

Probably not because $-\frac1{x^2}$ is also there but how do I do it then?

And can somebody explain me how to make the formulas and stuff look pretty for next times?

Tianlalu
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Danielvanheuven
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  • Do you mean $$e^{\frac{-1}{|x|}}$$? – Dr. Sonnhard Graubner Oct 14 '18 at 14:35
  • What you have called the "left limit" isn't. – Angina Seng Oct 14 '18 at 14:48
  • There doesn't seem to be any choice - $p$ is already fixed by continuity condition, differentiability can't bring a second condition. – orion Oct 14 '18 at 15:00
  • I don't understand what this quantity with $1/x^2$ is. But here is a tutorial on typing maths on this site https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – Calvin Khor Oct 14 '18 at 15:01
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    I think the OP is trying to investigate the derivative, not the original function itself. – orion Oct 14 '18 at 15:02
  • In addition, now that you know the limit of $e^{-1/|x|}$ from this previous question, you should accept one of the answers there https://math.stackexchange.com/questions/2955006/better-way-to-explain-lim-limits-x-to-0-e-frac-%E2%88%921x – Calvin Khor Oct 14 '18 at 15:04

1 Answers1

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A differentiable function must be continuous. Since $\lim_{x\to0}e^{-1/|x|}=0$, we get that $p=0$.

But is the function differentiable? Well, with $t=1/x$, $$ \lim_{x\to0^+}\frac{e^{-1/|x|}}{x}=\lim_{t\to\infty}{t}{e^{-t}}=0 $$ and with $t=-1/x$, $$ \lim_{x\to0^-}\frac{e^{-1/|x|}}{x}=\lim_{t\to\infty}{-t}{e^{-t}}=0 $$

egreg
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