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).
An idea. We take $\;x\;$ very close to zero, say $\;|x|&lt;10^{-4}\;$ :

$$x&gt;0:\;\;\frac1x-\frac1{x^2}=\frac{x-1}{x^2}&lt;\frac{-\frac12}{x^2}=-\frac1{2x^2}$$

and now you only have to show the rightmost expresion is unbounded below, which I think is pretty easy.
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
$$\lim\limits_{x \to 0}\left(\frac{1}{x} - \frac{1}{x^2}\right)=-\infty$$