Thread: Limits and Continuity

Hey all,

It seems easy but I dont seem to be getting the answer I am looking for

Let f(x)=(x+3) if x is less than -2
(Square root of(x+2) if x is greater than 1

I am looking to apply these too:

A) lim f(x) &
as x approches -2+

B) Lim f(x)
as x approaches -2-


Thanks alot, I am really happy to have found this site and I look forward to being a member.

A) lim f(x) &
as x approches -2+

Since in the interval -2 <= x <= 1, or [-2,1], the f(x) is not defined, then as x approaches -2 from the right, it is stopped from moving beyond x=1, where f(x) = f(1) = sqrt(1+2) = sqrt(3).
Therefore the limit is sqrt(3).

B) Lim f(x)
as x approaches -2-

As x approaches -2 from the left, the limit of f(x) is the intersection of f(x) = x+3 and the vertical line x = -2
So,
f(x) = x+3, and when x = -2, f(x) = -2 +3 = 1.
Therefore, the limit is 1.

Originally Posted by Svt-Sonic

Hey all,

It seems easy but I dont seem to be getting the answer I am looking for

Let f(x)=(x+3) if x is less than -2
(Square root of(x+2) if x is greater than 1

I am looking to apply these too:

A) lim f(x) &
as x approches -2+

This is undefined as f(x) is not defined arbitarily close to -2 from the right.

B) Lim f(x)
as x approaches -2-

this limit is +1, as f(x) is left continuous at -2.

RonL

Thanks alot guys!