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5 votes
Lim
t—>0 1-cos(2t) / sin^2 (11t)

1 Answer

2 votes

Answer:

2/121

Explanation:

Hello,

Hôpital's rule will help here.


\forall t \in \mathbb{R}^*\\\\f(t)=1-cos(2t)\text{ *** f is differentiable}\\f'(t)=2sin(2t)\\f''(t)=4cost(2t)\\f''(0)=4 \\ \\g(t)=sin^2(11t)\text{ *** g is differentiable}\\g'(t)=2sin(11t)* 11 * cos(11t)=2* 11 sin(11t)cos(11t)\\\\g''(t)=2* 11 (11cos^2(11t)-11sin^2(11t))\\g''(0)=2* 11^2

So the limit is 4/(2*121)=2/121

Thanks

User Alex Dn
by
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