Question

Determine the limit of the trigonometric function (if it exists).

lim t->0 (sin3t)/(2t)

Answer 1

Since this form is indeterminate you will have to use l'hospitals rule? Do you know how to use it?

Answer 2

never heard of it.

Answer 3

lol you need it!

Answer 4

No, you don't have to use l'Hospital's. I don't recommend it in limits without knowing derivatives.

Answer 5

We define \(u=\frac{1}{3}t\)
We know that as \(t\to0\), \(u\to 0\), so we can say:
\[
\lim_{t\to0}\left(\frac{\sin(3t)}{2t}\right)=\lim_{u\to0}\left(\frac{\sin(u)}{\frac{2}{3}u}\right)
\]We know, \(\lim_{x\to0}\frac{\sin(x)}{x}=1\) So:
\[
\lim_{u\to0}\left(\frac{\sin(u)}{\frac{2}{3}u}\right)=1\lim_{u\to0}\left(\frac{1}{\frac{2}{3}}\right)=\frac{3}{2}
\]Et voilá.
Another nice way to do it, is using asymptotic notation:
\[
\sin(x)\sim x, x\to0\]So, we substitute \(ax\) for \(\sin(ax)\) in the limit.

Answer 6

And, my bad on the \(\frac{2}{3}u\), it's supposed to be \(2u\).