Why does $e^{\frac10}\neq e^{\frac1{-0}}$?

Question

I was unable to explain why this fails? I asked to it many peers and they too can't. I faced this situation when solving a kind of integration problem.

Consider $x=-x$

Then $x=0$

That is, $0=-0$

Now consider,

$$e^{\frac1x}=e^{\frac1x}$$ $$e^{\frac1x}=e^{-\frac1{-x}}$$ $$e^{\frac10}=e^{-\frac1{-0}}$$ Now since $0=-0$

$$e^{\frac10}=e^{-\frac1{0}}$$ $$e^\infty=e^{-\infty}$$ $$\infty=0$$

But how can this happen?

UPDATED: $$\lim_{x\to 0+}\frac1x=\infty$$ $$\lim_{x\to 0-}\frac1x=-\infty$$

Then what is?

$$\lim_{x\to 0}e^\frac1x=?$$

Answer 1

The function $\frac1x$ is not continuous at $0$, and not only that it is not continuous, it has different limits from each side.

Consider the function: $f(x)=\begin{cases} -1 & x<0\\0 & x=0\\ 1 & x> 0\end{cases}$

It is clear that: $$\lim_{x\to0^-}f(x)=-1\neq 0=f(0)\neq\lim_{x\to0^+}f(x)=1$$

There's no reason to expect that a function discontinuous at $0$ will satisfy $\lim_{x\to0^-}f(x)=\lim_{x\to0^+}f(x)$.

And while some functions can be extended continuously to have a value at $+\infty$ or $-\infty$ (and this value itself may or may not be $\pm\infty$); not all functions can be extended in such way.

And even if you can extend it, there is no guarantee that $f(+\infty)=f(-\infty)$. Which is exactly the case for $f(x)=e^x$.

Answer 2

It's very simple:

$\frac10$ is not defined. So any equation that contains this term is undefined.

$\infty$ is not a number. It is the limit of an infinite process.

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For the second question about:
$\lim_{x\to 0}e^\frac1x=?$

As you did already proof by yourself, this term has two different limits. A left-side limit (by approaching from negative numbers and getting closer and closer to 0) and a right-side limit (by approaching from the other side), and those limits are different.

The point is, that $e^\frac10$ itself is undefined since $\frac10$ is undefined. You can get as close to $e^\frac1{x\to 0}$ as you want, and as long as $x\ne 0$ you will get well defined numbers as an result, but not for $x=0$. This case is undefined.

Answer 3

$\lim_{x\to 0-}e^\frac1x=e^{-\infty} = 0$

and

$\lim_{x\to 0+}e^\frac1x=e^{\infty} = \infty$

therefore the limit does not converge, therefore it does not exist.

graphing software can often be useful in testing limits like this.

earlier in your question you stated $1\over 0 $ $ = \infty$ this is somewhat of a naive approach to defining infinity that computer scientists sometimes use and is not mathematically rigours as in doing so they ignore the limit from the negative direction.