How to evaluate $\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$ I have to find the limit of following
$$\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$$
I have no idea how to start this one off.
How would I do it?
Do I just substitute the $0$? It doesn't look that easy and simple. The answer says it's negative infinity.
Please show me a solution without graphing(unless for better explanation).

As written, the limit is in the so-called “indeterminate form $\infty-\infty$”), so we want to rewrite it in another way to start with:
$$
\frac{1}{x}-\frac{1}{x^2}=\frac{x-1}{x^2}
$$
It's not restrictive to work under the assumption that $-1<x<1$; thus $|x|<1$ and $|x|^2<|x|$, that is to say
$$
\frac{1}{x^2}>\frac{1}{|x|}
$$
Since $\lim_{x\to0}(x-1)=-1$, we can restrict ourselves to an interval around $0$ where $x-1<-1/2$, so
$$
\frac{x-1}{x^2}<\frac{-1/2}{x^2}<-\frac{1}{2|x|}
$$
Since
$$
\lim_{x\to0}-\frac{1}{2|x|}=-\infty
$$
we are done.

However, this can be stated in greater generality; if you know that


*

*$\displaystyle\lim_{x\to a}f(x)=l>0$ (possibly $l=\infty$)

*$\displaystyle\lim_{x\to a}g(x)=0$

*$g(x)>0$ in a neighborhood of $a$ ($a$ excluded)


then
$$
\lim_{x\to a}\frac{f(x)}{g(x)}=\infty
$$
Note that the limit can also be for $x\to a^+$ or $x\to a^-$; changing into $l<0$ or $g(x)<0$ is easy with the “rule of signs”.
The proof is just the same as before: since $\lim_{x\to a}f(x)=l>0$, we can restrict ourselves to a (punctured) neighborhood of $a$ where $f(x)>k$ for some $k>0$. Then, since $\lim_{x\to a}g(x)=0$, for any $M>0$ we can choose $\delta>0$ so that, for $0<|x-a|<\delta$, $|g(x)-0|<k/M$. Thus, as we can also assume $g(x)>0$, $1/g(x)>M/k$ and
$$
\frac{f(x)}{g(x)}>k\frac{M}{k}=M
$$
This is exactly proving that $\lim_{x\to a}f(x)/g(x)=\infty$.
