Should we divide by 0 in limited question I had a huge argument between me and my doctor, It was a limit question, and we had $e$ to the power $1$ divided by $x$ and the $x$ was approaching zero. The doctor ended the question by dividing by zero and saying that $e$ to an infinite power is also infinite, and because we have one divided by the above formula it will equal one and I said that this is a mistake.
Here's how the doctor solved the question

$$\lim_{x \to 0}\frac{e^{-1/x} + 1}{e^{-1/x} - 1} = \lim_{x \to 0}\frac{\dfrac{e^{-1/x} + 1}{e^{-1/x}}}{\dfrac{e^{-1/x} - 1}{e^{-1/x}}} = \lim_{x \to 0}\frac{1 + \dfrac{1}{e^{-1/x}}}{1 - \dfrac{1}{e^{-1/x}}}$$
$$ = \lim_{x \to 0}\frac{1 + \dfrac{1}{e^{-1/0}}}{1 - \dfrac{1}{e^{-1/0}}} = \lim_{x \to 0}\frac{1 + \dfrac{1}{\infty}}{1 - \dfrac{1}{\infty}} = \fbox 1$$

I think that he is wrong and that we should not divide by zero, even if in this example the answer was right, but the steps were still wrong.
 A: Do I think he should've kicked you out from the class? Well, that depends. Your concern is legitimate, but it is all about how you raised it. So I'm not going to comment further; this is something that would be entirely out of my field of expertise, and is not my place to comment on. Raise that concern with your professor.

Now, the core of your question is whether we should consider this valid:
$$\lim_{x \to 0} e^{1/x} = e^{1/0} = \infty$$
The answer is "no," from a technical/rigor sense - but from an intuition standpoint it makes sense. It's common to think of $1/x$ "blowing up to infinity" as $x$ becomes smaller, but it's still invalid to declare the limit as such because
$$\lim_{x \to 0^+} \frac 1 x = \infty \;\;\;\;\; \text{but} \;\;\;\;\; \lim_{x \to 0^-} \frac 1 x = -\infty$$
Since the limits from each side are different, the limit as $x \to 0$ doesn't exist. More generally
$$\lim_{x \to c} f(x) \text{ exists if and only if both } \lim_{x \to c^+} f(x), \lim_{x \to c^-} f(x) \text{ do}$$
This is one of many reasons we don't even try to define $1/0 = \infty$. To claim $1/0 = \infty$ even in limits is only an appeal to intuition, often found in rudimentary calculus courses before proper rigor is introduced, and is by no means a hard rule.
So yes, your professor is wrong to do so, but not only is it an easy mistake to make, it's something that appeals to students' intuition. I suppose the jury is out as to whether appealing to intuition over rigor is a good idea or not, but I see why one might do that at least, probably even to avoid discussions like these. (Personally I'm in the camp of thinking it would be more important to deal with problems like these early on and handle things at least slightly more rigorously/properly, but I've yet to actually teach a class so ... who knows.)

In fact, this particular problem needs to stand out in your mind, because your professor's technique even leads to the wrong answer. This cements the point about $1/0 = \infty$ or $\lim_{x \to 0} 1/x = \infty$ being an intuitive tool at best, and not proper rigor. If we graph your function, we see

This is ignoring some other errors your professor made. Namely, the third limit should be
$$\lim_{x \to 0} \frac{1+e^{1/x}}{1 - e^{1/x}}$$
Do the simplification yourself to see why (he just divides the numerator and denominator by $e^{-1/x}$, or, equivalently, multiplies both by $e^{1/x}$). It should be clear that this limit is $-1$, not $+1$.
But anyways, check the graph above. Notice how, depending on which direction your approach is from, the limit as $x \to 0$ doesn't even exist, same as it doesn't exist for $1/x$ as $x \to 0$!
Thus,
$$\lim_{x \to 0} \frac{e^{-1/x} +1}{e^{-1/x} - 1}$$
doesn't even exist!
Why does this error arise? It is because your professor's idea uses
$$\lim_{x \to 0^+} \frac 1 x = \infty$$
with the sign on $0$ being important. After doing his trick, he is implicitly only doing a one-sided limit without realizing it. If he instead let it be $-\infty$ (which is equally as valid, i.e. not at all), he would get the limit to be $+1$ instead (ignoring the aforementioned other error).
