Question

Hello all, Can please someone help me with this question? with plenty or sufficient details please.Any kind of help would be greatly appreciated. Thank you.

Answer 1

a) We evaluate `f(2)` using the lower function because it includes `x = 2`.

`f(2) = 2/2 + 1 = 2`

b) For the limit from the positive side, you evaluate within the second function, because that is applicable for when `x` is to the right of `2`.

`lim_(x-> 2^+) = 2/2 + 1 = 2`

Vice versa for the left hand limit.

`lim_(x->2^-) = 3- 2 =1`

c) Since these two limits have different values they don't exist (`lim_(x-> 2^+) = lim_(x->2^-)` for `lim_(x->2)` to exist).

d) The left and right hand limits will have equal values as the second function (applicable for all `x ≥ 2`) is continuous.

`4/2 + 1 = 3`

Therefore, `lim_(x-> 4^+) = lim_(x-> 4^-) = lim_(x-> 4) = 3`

e) As explained in (d), yes `lim_(x-> 4)` exists.

Hopefully this helps!