Analysis: Finding limit with l'Hospital

  • #1
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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]
 
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Answers and Replies

  • #2
ehild
Homework Helper
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How did you use l'Hospital twice? Your method is legal. The result is correct.

ehild
 
  • #3
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How did you use l'Hospital twice? Your method is legal. The result is correct.

ehild
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.
 
  • #4
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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.
 
  • #5
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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.
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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