Understanding Continuity in Heaviside Equations?

[dud6913]

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.

Homework Equations

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]

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

 

[dud6913]

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]

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

 

[lanedance]

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

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

 

[dud6913]

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]

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;

 

[dud6913]

Thanks a lot!

 

[dud6913]

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

Ta