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Homework Help: Finding the limit of an expression involving three types of functions

  1. Sep 23, 2012 #1
    1. The problem statement, all variables and given/known data
    Find the limit (without using L'Hospital's rule)

    [tex]\lim_{x\to 0} \frac{ln(1 + sin^2x)}{e^{2x} - 1}[/tex]

    2. Relevant equations

    3. The attempt at a solution

    I tried various substitutions in order to rewrite the expression to a standard limit. Such as t = e^x, t = ln(x), t = sin(x) as well as using rewriting the trigonometric expression a few times. Since I suck I didn't get towards anything I recognized.

    I'd be grateful for any help sucking slightly less :)
  2. jcsd
  3. Sep 23, 2012 #2
    Can you convert it to the form 1^infinity?
  4. Sep 23, 2012 #3
    Not a clue :)

    I did however get help with a solution elsewhere.

    [tex]\lim_{x\to 0} \frac{ln(1 + sin^2x)}{e^{2x} - 1}[/tex]

    First multiply by [tex]\frac{sin^2x}{sin^2x}[/tex]

    Then [tex]\frac{ln(1 + sin^2x)}{sin^2x}[/tex] is on standard form and will approach one as x approaches zero.


    is multiplied by [tex]\frac{x^2}{x^2}[/tex]

    which gives the standard limit [tex]\frac{sin x}{x}[/tex] squared, which also approaches 1 as x approaches zero.

    [tex]\frac{x^2}{e^{2x} - 1}[/tex]

    Is multiplied by 2/2 which gives the remaining factors as standard limits.

    [tex]\frac{2x}{e^{2x} - 1}[/tex]



    Giving the final expression

    [tex]\lim_{x\to 0} \frac{ln(1 + sin^2x)}{sin^2x} \cdot \frac{sin^2x}{x^2} \cdot \frac{2x}{e^{2x} - 1} \cdot \frac{x}{2}[/tex]

    Since the first three factors approaches 1, and the fourth approaches 0, as x approaches 0 the limit will be 0.
  5. Sep 23, 2012 #4
    That's a good method. :smile:
    What i meant was writing the given expression as
    [tex]ln(1+\sin ^2x)^\frac{1}{e^{2x}-1}[/tex]
    Now, its of the form 1^infinity in the logartihm.
  6. Sep 23, 2012 #5
    Edit: Nevermind I misunderstood

    I don't really understand how that expression evaluates though.
  7. Sep 23, 2012 #6
    Here's my thinking:

    [itex]\frac{1}{e^{2x} - 1} \cdot ln(1 + sin^2 x) = ln(1 + sin^2 x)^{\frac{1}{e^{2x} -1}}[/itex]

    When x approaches 0

    [itex]ln(1 + sin^2 0)^{\frac{1}{e^{0} - 1}} = ln(1 + 0)^{\frac{1}{0}} = ln(1)^{∞} = ln(1) = 0[/itex]

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