How to evaluate $\lim_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)$
An idea. We take $\;x\;$ very close to zero, say $\;|x|<10^{-4}\;$ :
$$x>0:\;\;\frac1x-\frac1{x^2}=\frac{x-1}{x^2}<\frac{-\frac12}{x^2}=-\frac1{2x^2}$$
and now you only have to show the rightmost expresion is unbounded below, which I think is pretty easy.
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 $\lim_{x\to a}f(x)/g(x)=\infty$.