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avanturin [10]
3 years ago
8

Lim t—>0 1-cos(2t) / sin^2 (11t)

Mathematics
1 answer:
Nitella [24]3 years ago
6 0

Answer:

2/121

Step-by-step 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)\times 11 \times cos(11t)=2\times 11 sin(11t)cos(11t)\\\\g''(t)=2\times 11 (11cos^2(11t)-11sin^2(11t))\\g''(0)=2\times 11^2

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

Thanks

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Step-by-step explanation:

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