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Find the limits of the following functions

  1. Feb 4, 2008 #1
    1. The problem statement, all variables and given/known data

    Ok – two limit questions. Find the limit as x -> infinity or if no limit exists then prove so.

    i) (x + log(x^2))/(3x+2)
    ii) x/(1 + (x^2)(sin x)^2

    3. The attempt at a solution

    i) my first thought was that log(x^2)/x tends to 0 so the function tends to 1/3, but if this is the case do I need to prove that log(x^2)/x tends to 0?

    ii) well 0 <= (sin x)^2 <= 1 , im pretty stuck on this one.
  2. jcsd
  3. Feb 4, 2008 #2
    Does log here means ln? If so then by using L'hopital rule
    [tex]\lim_{x\to\infty}\frac{x+\ln x^2}{3x+2}=\lim_{x\to\infty}\frac{(x+\ln x^2)'}{(3x+2)'}=\lim_{x\to\infty}\frac{1+\frac{2x}{x^2}}{3}=\frac{1}{3}[/tex]

    [tex]\lim_{x\to\infty}\frac{x}{1+x^2\sin^2 x}=\lim_{x\to\infty}\frac{1}{2x\sin^2 x+x^2 2\sin x\cos x}[/tex]
    Last edited: Feb 4, 2008
  4. Feb 4, 2008 #3
    thanks - i) makes sense.

    so ii) -> 0 ?
  5. Feb 4, 2008 #4


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    Homework Helper

    Well, L'Hospital does not work for cases, in which, the limit does not exist. :wink:


    You can think like this: as x tends to infinity, sin(x) can take any value on the interval [-1, 1].

    If sin(x) = 0, your expression becomes: x.
    And if sin(x) = 1, then your expression will become: x / (1 + x2)

    So now, we'll choose 2 sequences (namely, xn, and x'n), both of which grow without bounds, and sin(xn) = 0, sin(x'n) = 1, for all n.

    We can choose xn = 2n(pi), and x'n = pi/2 + 2n(pi).
    The 2 sequences above satisfy all requirements above (you can check it yourself).

    What can you say about the 2 limits:

    [tex]\lim_{n \rightarrow \infty} \frac{x_n}{1 + x_n ^2 \sin ^ 2 (x_n)} , \quad n \in \mathbb{N} [/tex]


    [tex]\lim_{n \rightarrow \infty} \frac{x'_n}{1 + x'_n ^2 \sin ^ 2 (x'_n)} , \quad n \in \mathbb{N} [/tex]

    From there, what can you conclude about the limit:

    [tex]\lim_{x \rightarrow \infty} \frac{x}{1 + x ^2 \sin ^ 2 (x)}[/tex]?

    Can you go from here? :)
    Last edited: Feb 4, 2008
  6. Feb 5, 2008 #5
    yep - thanks
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