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Simple limit question

  1. Dec 12, 2004 #1

    mad

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    Hello, I tried doing this limit problem:

    lim (sinx)/x^3
    x-->0

    here's what I did:

    (lim sinx)/(lim x^3)

    which should give something like 0,000000000...1 / 0,00000000001
    since lim sinx (x->0) = 0 and lim x^3 = 0 too. When I make the graphic in maple, the answer is +infintiy, and the answer in my book too. Could someone clarify this for me plwase
    thanks a lot
     
  2. jcsd
  3. Dec 12, 2004 #2

    Pyrrhus

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    Homework Helper

    Do you know L'Hospital?

    [tex] \lim_{x \rightarrow 0} \frac{\sin x}{x^3} [/tex]

    [tex] \lim_{x \rightarrow 0} \frac{\cos x}{3x^2} = \frac{1}{0} = \infty[/tex]

    Alternatively you could use the fact that

    [tex] \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 [/tex]

    So we rewrite the first expression to

    [tex] \lim_{x \rightarrow 0} \frac{\sin x}{x} \frac{1}{x^2} [/tex]

    Applying limit laws:

    [tex] \lim_{x \rightarrow 0} \frac{\sin x}{x} \lim_{x \rightarrow 0} \frac{1}{x^2} [/tex]

    [tex] 1 \cdot \frac{1}{0} = \infty [/tex]

    Note: when you try to calculate the limit, you should try to get it without making a indeterminate form such as the one you got ([itex] \frac{0}{0} [/itex]).
     
    Last edited: Dec 12, 2004
  4. Dec 12, 2004 #3

    mad

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    Thank you very much for your help. I appreciate it :)
    What do you mean by L'hospital? In french we say L'hopital ;).. if its a theorem we haven't seen it yet.
    thanks again
     
  5. Dec 12, 2004 #4
    He means L'Hopital's rule. Mathworld can't seem to spell hopital :/, but read their page on L'Hopital's rule.
     
  6. Dec 12, 2004 #5

    Pyrrhus

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    Sorry, my native language is spanish :smile:, and i was taught L'Hospital like that, it appears like that on our books in spanish, and in the english books too, i've seen it too as L'Hospitel, oh well :smile:
     
  7. Dec 12, 2004 #6

    quasar987

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    Gold Member

    mad: The name of the theorem does not refer to an actual "hôpital" but as a dude whose last name was Guillaume François Antoine de L'Hospital (he was a Marquis in France I think).

    Also, be careful while using those properties of the limit such as the one Cyclovenom used, that is, since

    [tex] \lim_{x \rightarrow 0} \frac{\sin x}{x} = 1 [/tex]

    and

    [tex]\lim_{x \rightarrow 0} \frac{1}{x^2} = +\infty[/tex]

    therefor

    [tex] \lim_{x \rightarrow 0} \frac{\sin x}{x} \frac{1}{x^2} = +\infty[/tex]

    You have to watch out for the cases where the limits you "separated" have indeterminate forms (there are 7 of them). When you have a quotient, or product, or sum of limits that have this form, you cannot conclude to the value of limit. (as much as one would be tempted to conclude that [itex]\infty - \infty = 0[/itex] for exemple, we can not.)
     
    Last edited: Dec 12, 2004
  8. Dec 12, 2004 #7
    This page explains everything.
     
  9. Dec 12, 2004 #8

    mad

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    Thanks a lot guys!
     
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