Question

How do you evaluate the limit `s+2` as x approaches `0`?

Answer 1

See below.

Explanation:

If there is no error in the question and if `s` is independent of `x`, then

`lim_(xrarr0)(s+2) = s+2`

If the question should have been about `x+2` as `x rarr 0`, then we can reason as follows:

If `x` is getting closer and closer to `0`, then `x+2` is getting closer and closer to `0+2` which is `2`.

`lim_(xrarr0)(x+2) = 2`.

The result is the same as substituting `0` in for `x`. (Not every limit can be found by substitution, but this one can.)