Question

Evaluate the limit. $$ \lim _{t \rightarrow 0}\left(t^{2} \mathbf{i}+3 t \mathbf{j}+\frac{1-\cos t}{t} \mathbf{k}\right) $$

Short answer

The three-dimensional limit as \(t \rightarrow 0\) is \(0\mathbf{i}+0\mathbf{j}+0\mathbf{k}\), or the 0 vector.

Step-by-step solution

Solve for i

Evaluate the limit for \(t^2\). As t approaches 0, \(t^2\) also approaches 0. Therefore, the limit of the i component as t approaches 0 is 0.

Solve for j

Evaluate the limit for \(3t\). As t approaches 0, \(3t\) also approaches 0. Therefore, the limit of the j component as t approaches 0 is 0.

Solve for k

The k component of the limit is a bit challenging. It is in the form \(\frac{0}{0}\) as t approaches 0, which is indeterminate. However, by applying L'Hopital's rule (which states that the limit of the ratio of two functions is equal to the limit of their derivatives if the original limit is of the form \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)), this can be solved. The derivative of the numerator of the k component \(\frac{d}{dt}(1-cos(t)) = sin(t)\). The derivative of the denominator of the k component is simply 1. Therefore, the limit as t approaches 0 of the k component is \(\lim_{t \rightarrow 0}(sin(t))=0\).

Combine the limits

Now combine all the component limits to get the overall limit. Therefore, the three-dimensional limit as \(t \rightarrow 0\) is \(0\mathbf{i}+0\mathbf{j}+0\mathbf{k}\), or simply 0 vector. This means as \(t \rightarrow 0\), the vector also approaches the origin.