# Help with Heaviside equations

## Homework Statement

g(x) = [ x^2 + 2 if x<=1 & x + 2 if x>1,

I am asked to find the left and right limits, and whether the g(x) is continuous or not.

## The Attempt at a Solution

When I draw the two equations, I get a hyperbola and a line of gradient 1. They both share the same point(1), so I would say that the limit on the left and right is both 3? Now here is where it confuses me. All the answers given, have no relevance to my opinion.

lanedance
Homework Helper
i think you mean parabola and a line of gradient 1, and yes as they both tend to the same point, the function is conitnuous

Hi lanedance,

However, I am still unsure what the limits would be. I reckon that both left limit (x^2 + 2 if x<=1) and right limit( x + 2 if x>1), should be infinite.

Correct me if I am wrong.

Though, at x=1, the limit would be 3 right?

lanedance
Homework Helper
i think you're confusing limits at infinity and the left & right limit as you approach x=1

lanedance
Homework Helper
Though, at x=1, the limit would be 3 right?

yes as both the left and right limits exist and are the same

Also, i) the answers are that left limit is 1, and the right limit is 2, and that g(x) is continuous at x=1;
ii) that left limit is 2, and the right limit is 1, and that g(x) is not differentiable at x=1;
iii) that left limit is 1, and the right limit is 2, and that g(x) is not differentiable at x=1;
iv) that left limit is 1, and the right limit is 3, and that g(x) is not continuous at x=1;
v) none of the above

As i said before, i think that the limit for both of the functions would be at x=1, y=3, however, none of them give me that answer. Since you mentioned that the g(x) is continuous, i am assuming that the answer would be i).

Also, aren't both sides (left and right) meant to be continuous for g(x) to be continuous?
Left function includes 1, but right doesn't.

I am so confused....

lanedance
Homework Helper
sorry, aritmetic mistake
g(x) = [ x^2 + 2 if x<=1 & x + 2 if x>1]

clearly g(1) = 2 be defin

let x = 1-e, the left limit is as e>0 tends to 0
g(1-e) = (1+e)^2+2 = 3+e+e^2 --> 3, as e-->0

as x = 1+e, the right limit is as e>0 tends to 0
g(1+e) = 2+ (1+e) = 3 + e --> 3, as e-->0

so left limit is 3, right limit is 3, and that g(x) is continuous at x=1;

Thanks a lot!

I have also got the same answer after researching on the internet for a while.

Ta