Find lim t -->0 ( (tant/t) i - (3t/sint)j + (cos t + (t^2)/2 - 1) / t^2 ) k
We can evaluate the limit separately for each component of the vector.
For the i-component, we have:
lim t --> 0 (tant/t) = lim t --> 0 (tan t)/1 = 0
Therefore, the i-component approaches 0 as t approaches 0.
For the j-component, we have:
lim t --> 0 (3t/sint) = lim t --> 0 (3/sin t) * t = 3
Therefore, the j-component approaches 3 as t approaches 0.
For the k-component, we have:
lim t --> 0 (cos t + (t^2)/2 - 1) / t^2 = lim t --> 0 (cos t - 1)/t^2 + (1/2)
Using L'Hopital's rule, we can evaluate the first term as:
lim t --> 0 (-sin t)/2t = 0
Therefore, the entire k-component approaches 1/2 as t approaches 0.
Putting these results together, we get:
lim t --> 0 ( (tant/t) i - (3t/sint)j + (cos t + (t^2)/2 - 1) / t^2 ) k = 0i + 3j + (1/2)k
Therefore, the limit of the vector as t approaches 0 is <0, 3, 1/2>.