Question

How do you evaluate the limit `s^2/2+3s+11/2` as s approaches `0`?

Answer 1

The function is continuous in `s = 0`, so the limit equals the value.

Explanation:

For `s = 0`, you have

`lim_(s -> 0)(s^2/2 + 3 * s + 11/2) = 0^2/2 + 3 * 0 + 11/2 = 11/2`