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This is a problem from my introductory calculus homework.

\begin{equation} f(x) = \begin{cases} x^2\sin\frac{1}{x}, & x \neq 0,\\ 0, & x = 0. \end{cases} \end{equation}

Does $f'(0)$ exist?

If I approach the problem algebraically, I obtain the answer \begin{equation}f'(x)=\lim_{x \rightarrow 0}x\sin\frac{1}{x}\end{equation} This should be equal to 0 from the Squeeze Theorem. So, $f'(0)=0$.

However, something about this answer worries me. At class, the professor said to informally think of the tangent line at a point as the line that the curve begins to "look like" if we zoom around the point (or so I remember, I might be misquoting here). At the time this made a whole lot of sense to me. However, if I look at the problem from this angle, my geometric intuition goes beep, since I imagine the curve will never begin to look like a line :) So I wanted to know your thoughts on this one. Thank you in advance.

Michael Hardy
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Rashi
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    The amplitude of the oscillation shrinks faster than the frequency grows, so it will in the end look almost like a straight line. – Daniel Fischer Oct 06 '13 at 23:06
  • Yes, I kinda thought about that, but it somehow bugs me that I know that it is actually wavy. Maybe I'm just overthinking this. It is just that this case is somewhat different from the more obvious cases that I can think of, in which if we zoom in the line will become only very slightly curved. Maybe I just need more time to internalize all this. – Rashi Oct 06 '13 at 23:13
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    When you zoom in, the vertical dimension shrinks faster than the horizontal, and that makes the curves flatten out. Think of ripples moving towards the edge of a lake or a playground seesaw coming to rest. – Rob Arthan Oct 06 '13 at 23:27
  • Hahah, well, that is a beautiful way to think about it. Thanks to you two! – Rashi Oct 06 '13 at 23:28

1 Answers1

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By definition:

$$f'(0):=\lim_{x\to 0}\frac{f(x)-f(0)}x=\lim_{x\to 0}\frac{x^2\sin\frac1x}x=\lim_{x\to 0}x\sin\frac1x=0$$

since $\;\sin\frac1x\;$ is bounded

DonAntonio
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