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

1 Answer
Dec 13, 2016

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.)