Question

lim tan6t/sin2t as t->0

Answer 1

@uno we know
\[\lim_{t\to 0} \frac{\sin t}{t}\ \\ =\lim_{t \to 0} \frac{\tan t}{t}\ \ \ \ \ =1\]
here if we have t=2t or 6t
then also this is valid.
Can you use this to solve the question?

Answer 2

How did you get sin t/t?

Answer 3

@uno if you write the expansion of sin t
\[\sin t= t-t^3/3!+t^5/5!.....\]
so sin t/ t=
\[\sin t / t=1-t^2/3!+t^4/5!\]
if t=0 then
\[\sin t /t =1-0=1\]