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Homework Help: Limit Derivative Question

  1. Oct 6, 2006 #1

    This question has me confused which I kind of enjoy but I wondered what yall thought about it.

    [tex]F(x)=x^2*sin(\frac{3}{x})[/tex] when x ≠ 0
    F(x) = 0 when x=0

    What is the derative?
    Since, [tex]lim x^2*sin(\frac{3}{x})=0[/tex] so I know it is continous.

    If I were to guess the value of F'(x) I would guess 0 or undefined, but I don't how to show it and I was wondering if yall could help or give any pointers.
  2. jcsd
  3. Oct 6, 2006 #2


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    Here, you must be careful and pay particular attention to how the definition of the derivative is written!

    1. Find the derivative of F(x) at all points x DISTINCT from zero
    2. Find the derivative of F(x) AT x=0

    3. Take the limit as x goes to 0, if it exists, of what you found in 1.

    What does that tell you about F'(x) as a whole?
  4. Oct 6, 2006 #3
    Okay, using the limit definition:

    [tex]\lim_{h\rightarrow 0} \frac{f(x+h)-f(x)}{h}[/tex]

    [tex]\lim_{h\rightarrow 0} \frac{(x+h)^2*sin(\frac{3}{x+h})-(x)^2*sin(\frac{3}{x}}{h}[/tex]

    But at x=0:
    [tex]\lim_{h\rightarrow 0} \frac{(h)^2*sin(\frac{3}{h})-0}{h}=h*sin(\frac{3}{h})=3[/tex]

    But I am pretty sure I am making some mistake :frown:
  5. Oct 6, 2006 #4


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    Hmm..that was not exactly what I had in mind here (your answer is wrong, because it was complicated to start with)!

    Use the product rule to evaluate the derivative at an x DISTINCT from zero!
    Sorry for "misleading" you!

    But, you MUST use the limit definition in 2.!
  6. Oct 8, 2006 #5
    Step 1

    I know of two main ways to set up step two.
    [tex]lim_{x\rightarrow a} \frac{x^2sin(\frac{3}{x})-a^2sin(\frac{3}{a}){x-a}[/tex]

    [tex]lim_{h\rightarrow 0} \frac{(x+h)^2sin(\frac{3}{x+h})-x^2sin(\frac{3}{x}){x-a}[/tex]

    As far as actually evaluating the limit I am at a loss. However it occurred to me that I might have made a mistake before. Now I am not sure that limit of F(x) even goes to 0.

    The best reasoning I have come up with so far is that if we approach 0 from the derivative in step 1 from the left, then we multiply a really small negative number times the sin of a really big negative number. If we approach from the right we mulitply a really small positive number times the sin of a really big positive number. However I am not sure whether this is different or the same since sin is an odd function maybe the 2 negative sines cancel. Now I still don't know what to do.
  7. Oct 8, 2006 #6


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    There's only one way to set up the definition of the derivative AT x=0
    If it exists, we have:

    That is, F'(0)=0, and we see that the derivative F'(x) is DEFINED at x=0, but is DIS-continuous there!
  8. Oct 8, 2006 #7
    I am sorry. I know this must be frustrating. I don't see how you evaluated that limit. How would we evaluate:

    What I want to say is that
    -1< sin G <1
    and so then zero times any of those is zero. Is that logic correct?

    Since the limit is the same from both the right and the left it looks like it exists and is 0. It is discontinous because of plugging in values for solutoin of step 1.
  9. Oct 8, 2006 #8
    I think I understand much better now. Thanks for your help. The slope is 0 at x=0 from evaluating the limit.
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