limit \lim_{t \to 0}\frac{\sin 3t}{\tan2t} I am struggling with this question.

Question

limit

lim_{t \to 0} \frac{\sin 3 t}{\tan 2 t}

I am struggling with this question. I have attempted it, but I keep getting stuck on the following step:

lim_{t \to 0} \frac{1}{2} \frac{\sin 3 t \cos 2 t}{\sin t \cos t}

Answer 1

You started correctly. Now multiply both numerator and denominator by corresponding ts to get:

L = lim_{t \to 0} \frac{\sin {3 t}}{3 t} \cdot \frac{2 t}{\sin {2 t}} \cdot \frac{3}{2} \cdot \cos {2 t} = 1 \cdot 1 \cdot \frac{3}{2} \cdot 1 = \frac{3}{2} .

Note: It was used the remarkable limit:

lim_{x \to 0} \frac{\sin {x}}{x} = lim_{x \to 0} \frac{x}{\sin {x}} = 1

Answer 2

I suspect the denominateur is

\tan 2 t

. If I'm right, just use equivalents: near 0, we have

\sin u ~_{0} u, \tan u ~_{0} u

so that

\frac{\sin 3 x}{\tan 2 x} ~_{0} \frac{3 x}{2 x} = \frac{3}{2}

Answers (2)