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).
 A: 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.
A: 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 $\lim_{x\to a}f(x)/g(x)=\infty$.
A: $$\lim_{x\to 0}\frac{x-1}{x^2} = \lim_{x\to 0}\frac{-1}{x^2} = -\infty$$
A: For $f(x)=\frac{1}{x}-\frac{1}{x^2}$, if you want to know what $\lim_{x \rightarrow 0}f(0)$ is, try looking at $f(\frac{1}{10}),f(\frac{-3}{100})$ and selected similar values. That will start you off and give you an idea what the answer might be (or why the book says that the answer is what it is.) Once you have decided that the answer is $- \infty$ you might want to prove it. Graphing is, in some sense, a way to look at $f(x)$ for lots of values at the same time. 
A: A useful thing to do would be to make the substitution $u=\frac{1}x$. Then, this becomes
$$\lim_{u\rightarrow\infty}u-u^2$$
(or the analogous limit to $-\infty$) but $u$ grows much more slowly than $u^2$, the expression in the limit must decrease without bound - in particular, since $u^2>2u$ if $u>2$, we get that $u-u^2<-u$ if $u$ is at least $2$, so the limit is bounded above by $-u$ which goes to $-\infty$.
A: Here's another way to prove the following statement
$$\lim\limits_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)=-\infty$$
Which is equivalent to the following
$$\forall N\gt0,\exists\delta\gt 0:0\lt\left|x\right|\lt\delta\Rightarrow\frac{1}{x}-\frac{1}{x^2}\lt -N$$
So whenever $0\lt\left|x\right|\lt\delta$, we have
$$\frac{1}{x}-\frac{1}{x^2}=\frac{x-1}{x^2}\lt\frac{\delta-1}{x^2}$$
Here we can make $\left|x\right|$ arbitrarily small by presupposing a bound of $\frac12$. We can then use this bound to potentially get a smaller $\delta$ later. Assuming that $|x|\lt\frac12$, we have
$$x^2\lt\frac14\Rightarrow-\frac{1}{x^2}\lt-4$$
Which implies that
$$\frac{\delta-1}{x^2}=-\frac{1-\delta}{x^2}\lt -4(1-\delta)=-N$$
Now we can set $\delta$ to the smallest of these two bounds
$$\delta =\min\left(\frac12, 1-\frac{N}{4}\right)$$
Putting it all together
$$0\lt\left|x\right|\lt\delta\Rightarrow\frac{1}{x}-\frac{1}{x^2}\lt -4(1-\delta)=-N$$ 
Therefore
$$\lim\limits_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)=-\infty$$
