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Homework Help: L'Hospital's Rule problem

  1. Jan 6, 2008 #1
    Consider f(x) = x + sin(x)cos(x) and g(x) = e[tex]^{sin(x)}[/tex](x + sin(x)cos(x))

    Prove that lim[tex]_{x\rightarrow\infty}[/tex] f(x)/g(x) does not exist but that lim[tex]_{x\rightarrow\infty}[/tex] f '(x)/g '(x) = 0. Explain this.


    Well, it's pretty straightforward to see that lim f(x)/g(x) does not exist, since you end up with lim e[tex]^{-sin(x)}[/tex].

    But for the life of me I can't seem to work out how lim f '(x)/g '(x) = 0.

    According to L'Hospital's rule, a situation like this should never occur, but I think that it is also important to note that the equations do not meet the condition in L'Hospital's rule:

    If f(a) = 0 and g(a) = 0, then lim[tex]_{x\rightarrow\alpha}[/tex] f(x)/g(x) = lim[tex]_{x\rightarrow\alpha}[/tex] f '(x)/g '(x), provided that the limit on the right exists.

    In the problem, the limit on the right exists, but the limit on the left is not equal to the one on the right. Also, f(x) and g(x) do not equal zero when x is infinity.

    Can anyone help???
    Last edited: Jan 6, 2008
  2. jcsd
  3. Jan 6, 2008 #2
    L'Hopital's Rule says that if lim x->inf f(x)/g(x) is in an indeterminate form of 0/0 OR inf/inf then lim x->inf f(x)/g(x) = lim x->inf f'(x)/g'(x) provided the limit exists. I believe that both become infinitely large as they approach infinity
  4. Jan 6, 2008 #3
    Oh you're right, the problem is valid according to L'Hospital's law. And yes they do both become infinitely large as x approaches infinity.

    However - I still haven't worked out how lim f'(x)/g'(x) goes to zero....
  5. Jan 6, 2008 #4
    I'm not going to do out all the algebra, but basically here's what you get if you do it out

    [tex]f'(x) = 2\cos^2 x[/tex]
    [tex]g'(x) = e^{\sin x}[2\cos^2 x + \cos x(x + \sin x \cos x)][/tex]

    Dividing them, one of the cos will cancel, leaving

    [tex]\lim_{x\to\infty} \frac{2\cos x}{e^{\sin x}[2\cos x + x + \sin x \cos x]}[/tex]

    At this point, we see that the numerator is restricted to [-2, 2]

    In the denominator, the trig terms are restricted similarly and the exponential term is restricted to between 1/e and e. Because of this, as x approaches infinity, the linear x term will dominate; the others will be insignificant.

    The denominator will approach infinity while the numerator remains finite, so the limit goes to infinity.

    It is a case of

    [tex]\lim_{x \to\infty} \frac{some finite number}{approaching infinity} = 0[/tex]
  6. Jan 6, 2008 #5


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    Homework Helper

    it is also amusing to work out

    lim f''/g''
  7. Jan 6, 2008 #6
    I thought that f'(x) was 1+[cos(theta)]^2 - [sin(theta)]^2. But either way you're right about the limit! It's just weird because the limit on the bottom is approaching both positive and negative infinity, although I guess that doesn't matter. Thanks everyone!
  8. Jan 19, 2008 #7
    Hello again! It turns out that there is a problem with the explanation. I wrote on my paper that the limit of g'(x) as x goes to infinity is +- infinity, since g'(x) oscillates between positive and negative infinity. My professor wrote:

    "But cos(x) takes on zero value infinitely many times! This is not true that g'(x) goes to +- infinity as x goes to infinity, g'(x) vanishes infinitely many times as x goes to infinity and the limit of g'(x) does not exist!"

    So the reason that the problem does not follow L'Hospital's rule is because the rule does not allow g'(x) to oscillate. But that still leaves me back at square 1, in terms of proving that lim f'(x)/g'(x) = 0.

    Any suggestions?
    Last edited: Jan 19, 2008
  9. Jan 20, 2008 #8
    Ah, never mind! Figured it out!
  10. Jan 25, 2008 #9
    Chickendude was right about how lim f'(x)/g'(x) -> 0. The reason that this problem does not seem to follow L'Hospital's rule is because since g'(x) oscillates between positive and negative values, it will become zero infinitely many times as x approaches infinity, and L'Hospital's rule excludes functions in which g'(x) = 0 as x -> inf.
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