Is this answer to my math problem correct?

Question

1) show by means of an example that lim as x -->a [ f(x) + g(x) ] may exist even though neither lim as x -->a f(x) nor lim as x -->a g(x) exists

ANS: the limit of f(x) = 1/(x-a) has no limit as x goes to a. Neither does g(x) = -1/(x-a). However f(x) + g(x) => (1-1)/ (x-a) = 0, which is a constant so its limit exists.

Answer 1

Another very simple example:

f(x) = sin(1/x), x<>0
g(x) = 2 - sin(1/x), x <> 0

Neither f nor g have a limit at a = 0. But h(x) = f(x) + g(x) = 2 for every x <> 0 has limit 2 when x --> 0.

And your answer is right.

Answer 2

lim as x-->0 f(x)+g(x) = 1/(x-a) -1 /(x-a)
=0/(x-a)
as x-->a this limit does not exist, or takes the form 0/0.

Answer 3

it's impossible because

lim x--->a [f(x) + g(x)] = lim x--->a [f(x) ] + lim x--->a [g(x)]
so f(x) and g(x) must have a limit at (a)

Answer 4

Seems about right!