Analysis: Finding limit with l'Hospital

[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]

 

[ehild]

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

ehild

 

[Shackleford]

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.

 

[Mark44]

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.

 

[Shackleford]

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.

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.