Question

Find the limit of:

(t^3)/((tan2t)^3) as t->0

Answer 1

Using L'Hopital's rule? Or not? The former would be much simpler.\[\lim_{t\to0}\frac{t^3}{\tan^3{2t}}\]

Answer 2

Actually, either way is pretty simple. L'Hopital's is just less to type.

Answer 3

I do not know what L'Hopital's rule is. I only know the following:
\[\lim_{\theta \rightarrow 0} (\sin \theta)/\theta = 1\]

Answer 4

\[and \lim_{\theta \rightarrow 0} (\cos (\theta) -1)/\theta =1\]

Answer 5

I might have solved it.. Not sure.. My book doesn't give an answer

Answer 6

Basically, it's 8...

Answer 7

I got to go.. I will leave this up in case anyone wants to confirm my answer

Answer 8

Sorry, I was a bit busy. And yeah, you have to use the known limit of \(\frac{\sin\theta}{\theta}\) as \(\theta\) approaches 0.

Answer 9

\[\large\lim_{t\to0}\frac{t^3}{\tan^3{2t}}=\lim_{t\to0}\frac{t^3}{\frac{\sin^3{2t}}{\cos^3{2t}}}=\left[\lim_{t\to0}\frac{t\cos{2t}}{\sin{2t}}\right]^3\\\large=\left[\frac{1}{2}\lim_{t\to0}\frac{2t}{\sin{2t}}\cdot\lim_{t\to0}\cos{2t}\right]^3\]

Answer 10

Each inner limit is equal to 1, so you're left with \(\left(\frac{1}{2}\right)^3\not=8.\) But you were close!