The limit of →r(t), as t approaches c, is →L, expressed as means that given any ϵ > 0, there exists a δ > 0 such that for all t ≠ c, if |t - c| < δ, we have ∥→r(t) - →L∥ < ϵ.
Evaluate the limit of t t by plugging in 0 0 for t t. Evaluate the limit of 1 1 which is constant as t t approaches 0 0. Simplify the answer.
→r(t) = ⟨5cost, 3sint, 4sint⟩ on [0, 2π]. 43. →r(t) = ⟨t3, t2, t3⟩ on [0, 1]. 44. →r(t) = ⟨e - tcost, e - tsint⟩ on [0, 1]. 45. →r(t) = ⟨t, t2 √2, t3 3⟩ on [0, 3]. 46.
→r(t) is continuous at c if lim t → c→r(t) = →r(c). If →r(t) is continuous at all c in I, then →r(t) is continuous on I.
Move the limit inside the trig function because cosine is continuous.
L'Hospital's Rule states that the limit of a quotient of functions is equal to the limit of the quotient of their derivatives.
The derivative of cos ( t) cos ( t) with respect to t t is − sin ( t) - sin ( t).
At t = 1 each of the component functions is continuous. Therefore →r(t) is continuous at t = 1.
The theorem following the definition shows that in practice, taking limits of vector-valued functions is no more difficult than taking limits of real-valued functions.