Proving $\lim _{x \to 0^-} \frac{1}{x}=-\infty$ How do I calculate the limit:
$$\lim _{x \to 0^-} \frac{1}{x}$$
The answer is clearly $-\infty$, but how do I prove it? Its clear that as x approaches 0 from the right side, $x$, becomes infinitely small, but negative, making the limit go to $-\infty$, but how do I prove this mathematically?
 A: If you're not familiar with $\epsilon - \delta$ proofs, you can take numbers $x$ such that $x<0$ to show that $x \rightarrow -\infty$.   
ex: $1 \div -.0001 = -10000$. Keep dividing by smaller negative numbers to show that $x \rightarrow -\infty$.
A: Hint, to show $\forall a \lt 0, \exists \delta \gt 0 \text{ such that, } -\delta \lt x \lt 0,~~\frac{1}{x} \lt a$

$$\text{Let }\delta = -\frac{1}{a}$$

A: show x = -.0001 then x = -.00000001 then x = -.00000000000000001. Choose values that get closer and closer to 0 then graph that. That is the best way I could show it. Is this what you mean? Or am I making this too simple?
A: Notice $\displaystyle\frac{1}{x}$ is strictly decreasing on $(-1,0)$.
I claim that fo any positive integer $n$, if we choose $x$ close enough to $0$$, \displaystyle\frac{1}{x} < -n$. Hint: consider $x\in (-\frac{1}{n},0)$.
Do you see why it should follow that the limit is $-\infty$?
A: If we say 
$$
\lim_{y \to x} f(y) = f(x)
$$ 
Then, for 'x' approaching 0 from negative side,
$$
\lim_{x \to 0^-} \frac{1}{x} = \lim_{-\frac{1}{n} \to 0^-} \frac{1}{\frac{1}{n}}= -\lim_{-\frac{1}{n} \to 0^-}n = -\infty
$$
