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.
Best Answer
$\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.