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Evaluating limits

  1. May 27, 2006 #1
    I have trouble with this limit evaluation due to the fractions in it.

    [tex]\lim_{x \rightarrow 0} \left[3x^5cos\left(\frac{1}{x}\right)\right][/tex]

    I assumed that [tex]cos\left(\frac{1}{0}\right) = \infty[/tex] so that limit gives [tex](0 \cdot \infty)[/tex] and is an indeterminate form. As such, I rearrange so that I can use L'Hopital's rule:

    [tex]\lim_{x \rightarrow 0} \left[\frac{cos\left(\frac{1}{x}\right)}{\frac{1}{3x^5}}\right][/tex] this becomes [tex]\left(\frac{\infty}{\infty}\right)[/tex] and L'Hopital's rule applies.

    [tex]\Rightarrow \lim_{x \rightarrow 0} \left[\frac{\frac{sin}{x^2}\left(\frac{1}{x}\right)}{-\frac{5}{3x^4}}\right][/tex]
    Am I correct so far?
     
    Last edited: May 28, 2006
  2. jcsd
  3. May 27, 2006 #2

    VietDao29

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    Nah, it's not correct. In mathematics, you cannot assume things, you can just prove things.
    I'll give you a hint: What's the maximum, and minimum value of cos(x)? Another hint is that it's not infinity. :)
    Can you go from here? :)
     
  4. May 28, 2006 #3
    Maximum and minimum values of cos(x) is 1 and -1.

    If [tex]\lim_{x \rightarrow 0} cos\left(\frac{1}{x}\right) \neq \infty[/tex] then I cannot say that it is an indeterminate form and I cannot rearrange so that L'Hopital's can be used.
     
    Last edited: May 28, 2006
  5. May 28, 2006 #4

    VietDao29

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    No, you cannot rearrange that, since [tex]\cos \left( \frac{1}{x} \right)[/tex] is upper bounded by 1, and lower bounded by -1.
    Yes, this is correct.
    Now if x tends to 0, then 3x5 also tends to 0, and [tex]\cos \left( \frac{1}{x} \right)[/tex] oscillates between -1, and 1, right? So what does this limit tend to?
    Hint: And to prove that, you should use Squezze Theorem.
    Can you go from here? :)
     
  6. May 29, 2006 #5
    Oh ok. Here's my answer:

    Oh ok. Here's my answer:

    [tex]\lim_{x \rightarrow 0} \left[3x^5\cos\left(\frac{1}{x}\right)\right][/tex]
    [tex]-1 \leq \cos \left(\frac{1}{x}\right) \leq 1 (\times 3x^5)[/tex]
    [tex]\Rightarrow -3x^5 \leq 3x^5\cos\left(\frac{1}{x}\right) \leq 3x^5[/tex] since [tex]3x^5 > 0[/tex]
    [tex]\lim_{x \rightarrow 0} (-3x^5) = 0[/tex] and [tex]\lim_{x \rightarrow 0} (3x^5) = 0[/tex]
    [tex]\therefore \lim_{x \rightarrow 0} \left[3x^5\cos\left(\frac{1}{x}\right)\right] = 0[/tex] (Sandwich theorem)
     
    Last edited: May 29, 2006
  7. May 29, 2006 #6

    VietDao29

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    It's nearly correct. Congratulations. :)
    However if x < 0, then 3x5 < 0
    So you can split it into 2 cases:
    Case 1: x < 0
    [tex]3x ^ 5 \leq 3x ^ 5 \cos \left( \frac{1}{x} \right) \leq -3x ^ 5[/tex]
    So applying the Sandwich theorem here, we have:
    [tex]\lim_{x \rightarrow 0 ^ -} 3x ^ 5 = 0[/tex]
    And
    [tex]\lim_{x \rightarrow 0 ^ -} -3x ^ 5 = 0[/tex]
    So:
    [tex]\lim_{x \rightarrow 0 ^ -} 3x ^ 5 \cos \left( \frac{1}{x} \right) = 0[/tex]
    Case 2: x > 0 (It's x > 0, not x >= 0, since the function is not defined at x = 0)
    You can do almost exactly the same as case 1 (be careful, since there's a slight change in the inequality sign), and get the result:
    [tex]\lim_{x \rightarrow 0 ^ +} 3x ^ 5 \cos \left( \frac{1}{x} \right) = 0[/tex]
    So, from the 2 cases above, we have:
    [tex]\lim_{x \rightarrow 0} 3x ^ 5 \cos \left( \frac{1}{x} \right) = 0[/tex]
    :)
     
    Last edited: May 29, 2006
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