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