Analysis: Finding limit with l'Hospital

1. Nov 7, 2011

Shackleford

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. Nov 7, 2011

ehild

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

ehild

3. Nov 7, 2011

Shackleford

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. Nov 8, 2011

Staff: Mentor

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

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

I used a wellknown limit, $\lim_{t \to 0} \frac{sin(t)}{t} = 1$ and the fact that you can interchange the limit and function, provided the function is continuous at the limiting point.

5. Nov 8, 2011

Shackleford

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.