Q:

Differentiability: What if Left-hand and Right-hand Limit are Equal in $x$ but differ from $f(x)$?

Earlier today I asked this question: Derivative of Function with Cases: $f(x)=x^2\sin x^{-1}$ for $x\ne0$
Following the answers to my question I worked on my problem. Among other things I showed that the left hand limit of the derivative of the function from the question is equal to the left hand derivative. But that lead me to a new question:
A function is differentiable in $x$ iff $f'_+(x) = f'_-(x)$. But also a function isn't differentiable in $x$ if it is not continuous in $x$. What if both things are true?
The function $$f(x) = \begin{cases}x^2\sin x^{-1} & \text{if } x \neq 0\\ 10 & \text{else }\end{cases}$$ for instance. Here the right hand derivative is equal to the left hand derivative:
from the left:
\begin{align*}
  \lim_{x \to 0^-} \frac{f(x) - f(0)}{x - 0} & = \lim_{x \to 0^-} \frac{x^2 \sin \left( \frac{1}{x} \right) - {0^2 \sin \left( \frac{1}{0} \right)}}{x - 0}\\
                                         & = \lim_{x \to 0^-} \frac{{x} \left(x\sin \left( \frac{1}{x} \right) \right)}{{x}}\\
                                         & = \lim_{x \to 0^-} \frac{{x\sin \left( \frac{1}{x} \right)}}{1}\\
                                         & = 0
\end{align*}
from the right:
\begin{align*}
  \lim_{x \to 0^+} \frac{f(x) - f(0)}{x - 0} & = \lim_{x \to 0^+} \frac{x^2 \sin \left( \frac{1}{x} \right) - {0^2 \sin \left( \frac{1}{0} \right)}}{x - 0}\\
                                         & = \lim_{x \to 0^+} \frac{{x} \left(x\sin \left( \frac{1}{x} \right) \right)}{{x}}\\
                                         & = \lim_{x \to 0^+} \frac{{x\sin \left( \frac{1}{x} \right)}}{1}\\
                                         & = 0
\end{align*}
So, as the left and right hand limits are the same $f$ must be differentiable in $0$, right? But it is obviously not continuous at $0$, as $f(0) = 10$ but $\lim_{x \to 0} f(x) = 0$, so then it must not be differentiable... What am I missing here? Is the left-right-hand limit equality just a necessary rather than a sufficient precondition for differentiability?

A:

1.There is no  real number $\frac {1}{0}.$ Any computations using it are false or meaningless.

If $f(x)=x^2 \sin 1/x$ for $x>0$ and $f(x)=10$ for $x\leq 0$ then for $x>0$ we have $$\frac {f(x)-f(0)}{x-0}=\frac {x^2\sin 1/x -10}{x}=x(\sin 1/x) -10/x $$which has no limit as $x\to 0.$ So $f$ has no "upper" derivative at $0.$

