How do we calculate the Right and Left Hand Limit of 1/x? I am confused regarding one sided limits and how to calculate it.
For Example:
$$\lim_{x\to 0}\frac{1}{x}\quad\text{does not exist}$$
How can I validate that $\lim\limits_{x\to 0^+}\frac{1}{x}$ or $\lim\limits_{x\to 0^-}\frac{1}{x}$ exists?
I am pretty certain that the limits do exist because if we take a positive value which is small $0.0000000\dots1$ we will get a very big positive limit value.  And the same for the negative value.  I know that in this case RHL$\ne$LHL.
I hope somebody can help me figure this one out. Thank you.
Sorry could not write down the equations, hope my explanation is clear enough.
 A: $\mathbf{Definition} $: 
$$ \boxed{ \lim_{x \to a^+ } f(x) = \infty } $$
means that for all $\alpha > 0$, there exists $\delta > 0$ such that if $ 0<x -a < \delta$, then $f(x) > \alpha$
$\mathbf{Example} $:
$$ \lim_{x \to 0^+} \frac{1}{x} = \infty $$
We use the definition to establish this fact. In other words, for a given $\alpha > 0$, we need to find a $\delta > 0$  such that if $0< x < \delta $, then $\frac{1}{x} > \alpha $
Notice $$ \frac{1}{x} > \alpha \iff \frac{1}{\alpha} > x$$
Hence, if we select $\delta = \frac{1}{\alpha } $, we will achieve our desired result.
A: To say that $\frac1x \to \infty$ as $x \to 0^+$ (the plus means goes to $0$ from the positive direction) is to say that for any large $N > 0$ I can choose a small $\epsilon > 0$ (whose choice depends on $N$) such that $0 < x < \epsilon$ implies $\frac1x > N$.  This, of course, is achieved by choosing $\epsilon = \frac1N$.
I'll leave the negative side to you, it's almost identical.
A: From a bit of intuition, you can see:
$$\lim_{x \to 0^+}(1/x) = \infty$$
$$\lim_{x \to 0^-}(1/x) = -\infty$$
Of course, this is because as the value for $x$ gets infinitesimally small, the fraction blows up.  To actually prove these limits in a rigorous fashion, however, you'd need to do the following:
For the first limit: Given any $n \in \mathbb{R}$, show that there exists an $\epsilon > 0$ such that, for all $0<x<\epsilon$, we'll have $1/x > n$.
And you'd proceed likewise for the second.
