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Analysis: Finding limit with l'Hospital

  1. Nov 7, 2011 #1
    I'm not sure what I did with this expression is right.

    I looked at the argument as x tends to 0. I used l'Hospital twice and found that the argument tends to pi/3. Of course, tan(pi/3) = sqrt3.

    http://i111.photobucket.com/albums/n149/camarolt4z28/Untitled-3.png [Broken]
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 7, 2011 #2

    ehild

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    Homework Helper
    Gold Member

    How did you use l'Hospital twice? Your method is legal. The result is correct.

    ehild
     
  4. Nov 7, 2011 #3
    Oops. You're right. I just used it once.

    I plugged this into my TI-89 Ti, and it didn't agree with my analytical result. I suppose it's not very reliable as it didn't agree for a couple of other problems, too.
     
  5. Nov 8, 2011 #4

    Mark44

    Staff: Mentor

    It's worth noting that you don't need to use L'Hopital's Rule at all for this problem.

    [tex]\lim_{x \to 0^+}\tan\left( \frac{sin 2\pi x}{6x}\right)[/tex]
    [tex]=\tan (\lim_{x \to 0^+}\left( \frac{sin 2\pi x \cdot 2\pi x}{2\pi x \cdot 6x}\right))[/tex]
    [tex]=\tan (\lim_{x \to 0^+}\left( \frac{sin 2\pi x}{2\pi x} \cdot \frac{2\pi x}{6x}\right))[/tex]
    [tex]=\tan (\frac{\pi}{3}) = \sqrt{3}[/tex]

    I used a wellknown limit, [itex]\lim_{t \to 0} \frac{sin(t)}{t} = 1[/itex] and the fact that you can interchange the limit and function, provided the function is continuous at the limiting point.
     
  6. Nov 8, 2011 #5
    I like that method better. Ah. I didn't know that. I haven't read the section yet, but I'm sure it mentions being able to interchange the limit and function.
     
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