Differentiability of $x^2$ I tried checking differentiability of $x^{2}$ and on its right hand side, rate of change of $y$
with respect to $x$ was $0^{+}$ and on left hand side was $0^-$.
What does this $0$ imply? And how can we say it is differentiable at $x = 0$ since they are not same?
Also can anyone please help me comparing it with $|x|$ since its right hand derivate is $1$ and left hand derivative is $-1$. So, does it imply that if I increase value of $x$ by $1$, $y$ increases by $1$ on right hand side?
Here is the
$$f'(0^+)=\lim_{h\to 0^+}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^{+}}\frac{h^{2}}{h}=\lim_{h\to 0^{+}}h=0^{+}$$
$$f'(0^-)=\lim_{h\to 0^-}\frac{f(0+h)-f(0)}{h}=\lim_{h\to 0^{-}}\frac{h^{2}}{h}=0^{-}$$

 A: This is the problem with the ‘handwave’ definition of a limit that is taught in first year calculus courses.
Supposing $f(y)=z$, then for example $\lim_{x \rightarrow y} f(x) = z$ is shorthand for writing: for any $\epsilon>0$ there is a $\delta>0$ such that for all $x$ such that $x \in (y-\delta, y+\delta)$ we have $f(x) \in (z-\epsilon, z+\epsilon)$. In other words, as $x$ gets close to $y$ from any direction, then $f(x)$ gets close to $z$.
On the other hand, $\lim_{x \rightarrow y^+} f(x) = z$ is shorthand for writing: for any $\epsilon>0$ there is a $\delta>0$ such that for all $x$ such that $x \in (y, y+\delta)$ we have $f(x) \in (z-\epsilon, z+\epsilon)$. In other words, as $x$ gets close to $y$ from the right, then $f(x)$ gets close to $z$.
The meaning of $\lim_{x \rightarrow y^-} f(x) = z$ is similar to $\lim_{x \rightarrow y^+} f(x) = z$. It is then clear that $\lim_{x \rightarrow y^-} f(x) = z$ and $\lim_{x \rightarrow y^+} f(x) = z$ together is equivalent to $\lim_{x \rightarrow y} f(x) = z$.
