1
$\begingroup$

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?

  • 4
    What you're missing is that $f(x) - f(0) = f(x) - 10$. – 2017-01-14
  • 0
    if the left hand limit and the right hand limit are equal but the function is not defined at the point the function then we dont care about the point and i belive we can actually take the derivative of the funtction even at that point. – 2017-01-14
  • 1
    $f(0)$ is $10$ and not $(0^2 \sin \left( \frac{1}{0} \right))$ according to your definition of $f$. Besides that the latter expressiond does not make sense because of $\frac{1}{0}$. – 2017-01-14
  • 0
    @miracle173 you can bound sin by $x^2 $ and show that it dominates sin for all x clearly $x^2$ goes to 0 so sin(1/0) also goes to 0. – 2017-01-14
  • 0
    @Faust7 No, sin(1/0) does not go to 0, it is not defined. sin(1/0) isn't even an expression that contains a limit. – 2017-01-14
  • 0
    @miracle173 But whatever it is, it get's annihilated by 0, does it not? – 2017-01-14
  • 1
    @user3578468 It's not relevant. The definition of $f$ says that when $x = 0$, $f(x) = 10$, no matter what expression $f$ has otherwise. The _limit_ of $f$ as $x \to 0$ is $0$, because $x^2$ pushes it to $0$, but the _value_ at $0$ is $10$. – 2017-01-14
  • 0
    @user3578468 No, it simply is not defined and so the expression does not make sense. – 2017-01-14
  • 0
    @miracle173 sorry but your wrong not only is it contious at 0 its much more than that its $C^{1}$ – 2017-01-14
  • 0
    @user3578468 even with the piecewise definition give by you on every neighbourhood of 0 your function approach's 0 from $0^{+}$ and from$ 0^{-}$ so it is continuous the value at 0 is a removable point of discontinuity. if you look up epsilon delta proofs you can show this. – 2017-01-14
  • 0
    @Faust7 Does that mean now that when taking the limit from one side you need to put in the defined case ($f(0) = 10)$ for $f(0)$ -- as the others say -- or the value the function has as it approaches 0 ($x^2 \sin(x)$) as I did? And where can I find a proof or a motivation for whatever is correct? Putting in 10 I find weird, as then I get the wrong limit, as I see it. – 2017-01-14
  • 0
    @Arthur Ok, but I am investigating the limit... so, why would I put in the value that is defined for 0 rather than the limit behaviour? – 2017-01-14
  • 3
    You're investigating the _derivative_ at $0$, and part of that involves evaluating $f$ _exactly_ at $0$. – 2017-01-14
  • 0
    @Arthur Ok, are there certain terms I can search for when I want to see a proof on that to convince myself of this? – 2017-01-14
  • 2
    You can look at the definition of $f'(0)$, which is $\lim_{h \to 0}\frac{f(h) - f(0)}{h}$. You see the $f(0)$ in there? That's not a limit, that's _actually_ $f(0)$. How can you convince yourself of that? Because that's what it says. If we had meant something else, we would've _written_ something else. – 2017-01-14
  • 0
    @Faust7 no, you are wrong. I appears to me you are talking about the expression $x^2 \sin \left( \frac{1}{x} \right)$ and its continous continuation at $0$. This is in $C_1$. But actually we are talking about the expression $0^2 \sin \left( \frac{1}{0} \right)$. – 2017-01-14

1 Answers 1

3

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

  1. 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 $0.$
  • 0
    What do you mean by that? $\lim_{x \to 0} \frac{x \sin \frac{1}{x} - 10}{x} = \lim_{x \to 0} \frac{x \sin \frac{1}{x}}{x} - \lim_{x \to 0} \frac{10}{x} = 0 -\infty$ – 2017-01-14
  • 2
    Infinity is not a real number either. When we say that something goes to infinity, it is a euphemism for a longer rigorous statement. And $ \lim_{x\to \infty}f(x)=y$ is a short form for a longer rigorous statement that is meaningful only when $y\in \mathbb R.$ – 2017-01-14