Question

Hello, uh could you help me solve lim as u approaches 0 of (u^2cosu)/(sin2u)

Answer 1

\[\lim_{u\to0}\frac{u^2\cos u}{\sin2u}~~?\]

Answer 2

yes plz

Answer 3

Recall the identity, \(\sin2u=2\sin u\cos u\).
\[\begin{align*}\lim_{u\to0}\frac{u^2\cos u}{\sin2u}&=\lim_{u\to0}\frac{u^2\cos u}{2\sin u\cos u}\\\\
&=\frac{1}{2}\lim_{u\to0}\frac{u^2}{\sin u}\\\\
&=\frac{1}{2}\left(\lim_{u\to0}\frac{u}{\sin u}\right)\left(\lim_{u\to0}u\right)\end{align*}\]

Answer 4

wait but what is the solution? is there no limit, also could you shoe with the limits approaching infinity and negative infinity please

Answer 5

\[\begin{align*}\lim_{u\to0}\frac{u^2\cos u}{\sin2u}&=\lim_{u\to0}\frac{u^2\cos u}{2\sin u\cos u}\\\\
&=\frac{1}{2}\lim_{u\to0}\frac{u^2}{\sin u}\\\\
&=\frac{1}{2}\left(\lim_{u\to0}\frac{u}{\sin u}\right)\underbrace{\left(\lim_{u\to0}u\right)}_{\Large \text{Here's the hint.}}\end{align*}\]

Answer 6

Put more effort in reading SithsAndGiggles's reply, please.

Answer 7

thanks guys