Question

How do I solve for C and evaluate the limit in the problem attached?

Answer 1

if that limit is going to exist, since the denominator is 0 the numerator must be 0 at 13 as well

Answer 2

so
\[t^2-20t+C=(t-13)(t+\text{something})\]

Answer 3

and since your "middle term" is -20 and you have -13t all ready from multiplying this must mean that "something" is -7 so your numerator has to be
\[(t-13)(t-7)=t^2-20t+91\] if my arithmetic is correct.

Answer 4

then in order to take the limit you write
\[\frac{(t-13)(t-7)}{t-13}=t-7\] and
\[\lim_{t \rightarrow 13}t-7=13-7=5\]