Question

solve lim x-->0 sin^2(8t)/t

Answer 1

with or without l'Hospital ?

Answer 2

here's without l'hospital:\[\lim_{t \rightarrow 0}\frac{\sin^2(8t)}t=8\lim_{t \rightarrow 0}\sin(8t)\frac{\sin(8t)}{8t}\]now evaluate

here's with l'hospital:\[\lim_{t \rightarrow 0}\frac{\sin^2(8t)}t=\lim_{t \rightarrow 0}\frac{16\sin(8t)\cos(8t)}1\]now evaluate