Why is this limit positive infinity and not negative infinity? I've come across the limit shown below, but when I plug the value in, it becomes -1/0, which, to my understanding, should result in $ -\infty $. My question is why does it actually result in $\infty$ (I've checked)?
$$ \lim_{x \to 0}-\frac{e^x}{x-xe^x} $$
 A: To handle limits that approach $\infty$ or $-\infty$, there are two main things to keep in mind:

*

*Don't try to "plug in the value". Instead, once you realize the limit is going toward $\pm \infty$, you should focus on what happens when you plug in something very close to the target value. Figure out whether each term will be positive vs negative, and combine those to get the answer.

*Remember that there's very often a difference depending on what side you approach from. So be careful to specify whether you're trying to compute $\lim_{x \to 0 ^-}(\cdots)$ vs $\lim_{x \to 0 ^+} (\cdots)$.

Combining those pieces of advice, I'll show how to handle the example $\lim_{x \to 0 ^-} \frac{-e^x}{x-x e^x}$. The very first step is still to try to plug in 0 and get $\frac {-1} 0$, which tells us that the limit will be $\pm \infty$. Now we need to work harder to find out which one it will be.
Following the advice from (1) and (2) above, I'll focus on what happens when we plug in a value of $x$ that's very close to $0$ but on the negative side. You could imagine $x = -0.0000001$ as you think about this.
In the numerator, we have $- e^x$. We know $e^x > 0$ for any real $x$, so the numerator will be negative.
The denominator is $(x)(1-e^x)$. The first term is $x$ which is negative. Is the second term positive or negative? Well, $e^x$ is an increasing function and $e^0 = 1$, so $e^{\text{some slightly negative number}}$ will be slightly less than 1, so $1-e^x > 0$.
Putting things together, and simplifying by writing only the signs of each term, we have $$\frac{(-e^x)}{(x)(1-e^x)} = \frac{(-)}{(-)(+)} = (+)$$ so the overall expression is positive when $x$ is a small negative number. So we conclude $\lim_{x \to 0 ^-} \frac{-e^x}{x-x e^x} = +\infty$.
Notice this only found the limit as $x \to 0^-$. If you wanted the other side $\lim_{x \to 0^+}$ then you could go through basically the same approach but for small positive $x$ values instead. Once you get the hang of the method, you can often pretty much write out the computation once and then quickly read off both sides' limits from the result, but when you're getting started I'd really recommend just writing it out two separate times to handle each side.
