INSTRUCTOR: Calculate the limit as t approaches negative 2 of the vector valued function r of t for the function r of t equal to the square root of t squared minus 3t minus 1 quantity i plus, parentheses, 4t plus 3, close parentheses, j plus sine of numerator, parentheses, t plus 1, close parentheses, pi, with the denominator of 2, k.

Now, because each of the component functions are continuous on their entire domains-- on all of each of their domains, that is-- we can find this limit, the limit as t approaches negative 2 of r of t, by simply evaluating at each of these values since negative 2 is in each of their domains.

So for the first function, we have the square root of t squared minus 3t minus 1. So if we evaluate that at negative 2, that would be negative 2 squared, which is 4, minus 3 times negative 2-- that would be plus 6-- and minus 1, which would be 9, so the square root of 9, which is 3.

So our limit is 3i plus-- and 4t plus 3, evaluating that at negative 2 would give us a negative 5. So that would actually be 3i minus 5j.

And then, finally, evaluating sine of numerator t plus 1 quantity times pi over 2. t plus 1, when t is negative 2, that would be negative 1, so negative pi over 2. Sine of negative pi over 2 is negative 1. So this is minus k. So the limit as t approaches negative 2 of r of t equals 3i minus 5j minus k.