91Ó°ÊÓ

Consider the component that is in the i direction: \( \sqrt{t} \). The limit of this function as \( t \) approaches 1 can be found directly by substituting 1 into the function, which gives \( \sqrt{1} = 1 \). Therefore, the limit of the i component as \( t \) approaches 1 is 1.

Now, consider the j component of the function: \( \frac{\ln t}{t^{2}-1} \). The limit of this function as \( t \) approaches 1 is a 0/0 indeterminacy. A good technique to solve this type of indeterminacy is using L'Hopital's Rule which states that the limit of the ratio of two functions is equal to the limit of the ratio of their derivatives if the original limit results in an indeterminacy. Applying L'Hopital's Rule, the derivative of the numerator \( \ln t \) is \( \frac{1}{t} \) and the derivative of the denominator \( (t^{2}-1) \) is \( 2t \). Applying these derivatives to the limit gives the new limit \( \lim _{t \rightarrow 1}\left(\frac{\frac{1}{t}}{2t} \right) = \lim _{t \rightarrow 1}\left(\frac{1}{2t^{2}} \right) \). At \( t = 1 \), the new limit is \( \frac{1}{2} \). Therefore, the limit of the j component as \( t \) approaches 1 is \( \frac{1}{2} \).

Lastly, consider the component of the function in the k direction: \( 2t^{2} \). The limit of this function as \( t \) approaches 1 can be found directly by substituing 1 into the function yielding \( 2(1^{2}) = 2 \). Therefore, the limit of the k component as \( t \) approaches 1 is 2.

The final step is to combine the limits of the i, j and k components to form the overall limit of the vector function. This yields \( \lim _{t \rightarrow 1}\left(\sqrt{t} \mathbf{i}+\frac{\ln t}{t^{2}-1} \mathbf{j}+2 t^{2} \mathbf{k}\right) = (1i + \frac{1}{2}j + 2k) \)