Understanding Limits: Does the Sum of Limits Always Equal the Limit of the Sum?

[PhysicsAnonnn]

Homework Statement

Assume that the limit, as x approaches 0, of f(x)+g(x) exists. Must the limit be f(0)+g(0)? If it's true, explain. If it's false, give an example.


I am failing to understand this problem. How do I check if it's true or false?

 

[Mark44]

Homework Statement

Assume that the limit, as x approaches 0, of f(x)+g(x) exists. If it's true, explain. If it's false, give an example.


I am failing to understand this problem. How do I check if it's true or false?

I don't understand it, either. Are you sure you have stated the problem verbatim? It seems to me that you are missing something important in the problem statement.

 

[PhysicsAnonnn]

I don't understand it, either. Are you sure you have stated the problem verbatim? It seems to me that you are missing something important in the problem statement.

Sorry, I forgot to include the other part of the question. Must the limit be f(0)+g(0)?

 

[HallsofIvy]

In other words, "If $\lim_{x\to 0} f(x)+ g(x)$ exists, must it be equal to f(0)+ g(0)"?

Saying that $\lim_{x\to a} F(x)= F(a)$ is essentially saying that F is continuous at x= a, isn't it? Is there any part of your problem, that you have not told us, that says f and g are continuous at x= 0?

Suppose f(x)= 5 if x is not 0, f(0)= 100, g(x)= 4 if x is not 0, g(0)= 100. What is f(x)+ g(x)? What is $\lim_{x\to 0} f(x)+ g(x)$? What is f(0)+ g(0)?

 

[Mark44]

To add to what HallsOfIvy said about continuity, consider f(x) = (x² - 4)/(x - 2)

It can be shown that $$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$$
so this limit exists. Does that mean f(2) exists?

 

[PhysicsAnonnn]

In other words, "If $\lim_{x\to 0} f(x)+ g(x)$ exists, must it be equal to f(0)+ g(0)"?

Saying that $\lim_{x\to a} F(x)= F(a)$ is essentially saying that F is continuous at x= a, isn't it? Is there any part of your problem, that you have not told us, that says f and g are continuous at x= 0?

Suppose f(x)= 5 if x is not 0, f(0)= 100, g(x)= 4 if x is not 0, g(0)= 100. What is f(x)+ g(x)? What is $\lim_{x\to 0} f(x)+ g(x)$? What is f(0)+ g(0)?

f(x) + g(x) would be equal to 9 while f(0) + g(0) would be equal to 200. The only thing that matters is how f(x) + g(x) is defined near 0, but never considering x = 0. Is that correct?

 

[PhysicsAnonnn]

To add to what HallsOfIvy said about continuity, consider f(x) = (x² - 4)/(x - 2)

It can be shown that $$\lim_{x \to 2} f(x) = \lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$$
so this limit exists. Does that mean f(2) exists?

No, f(2) is undefined; therefore, $$\lim_{x \to 2} \frac{x^2 - 4}{x - 2} = 4$$ cannot be equal to f(2). The limit does not care about x = 2, it only cares about how it's defined near 2. Am I correct?

 

[Mark44]

Yes. Now can you answer your original problem?

 

[PhysicsAnonnn]

Yes, thank you Mark44 and HallsofIvy.
The easiest way to prove that the original problem is false is to substitute an equation for either f(x) or g(x) where f(0) or g(0) is undefined. So, I could use $$\lim_{x \to 0} {f(x)} + \lim_{x \to 0} {g(x)} = \lim_{x \to 0} \frac{1}{x^2} + \lim_{x \to 0} {x} = \infty$$
Since f(0) is undefined for f(x), the original is problem is false.
Is that correct?

 

[Mark44]

No, that's not a good counterexample. You want lim (x -> 0) (f(x) + g(x)) to exist, but f(0) + g(0) to be undefined. Think about what was said in this thread about continuity.

 

[Char. Limit]

No, that's not a good counterexample. You want lim (x -> 0) (f(x) + g(x)) to exist, but f(0) + g(0) to be undefined. Think about what was said in this thread about continuity.

But that limit does exist, doesn't it? It approaches the same value (positive infinity) from both sides of the limit, after all...

 

[PhysicsAnonnn]

I'm fine with anything other than limits as x approaches 0. Can you give me a counterexample for the original problem that works? For limits as x approaches 0, the only thing I can think of to make f(0) or g(0) undefined is an infinity function.

 

[Mark44]

But that limit does exist, doesn't it? It approaches the same value (positive infinity) from both sides of the limit, after all...

Only in the sense that infinity exists. What I was hoping for was something where the limit was a finite number, but the two functions were undefined.

 

[Mark44]

I'm fine with anything other than limits as x approaches 0. Can you give me a counterexample for the original problem that works? For limits as x approaches 0, the only thing I can think of to make f(0) or g(0) undefined is an infinity function.

f(x) = x²/x has a removable discontinuity at x = 0.

 

[PhysicsAnonnn]

f(x) = x²/x has a removable discontinuity at x = 0.

$$\lim_{x \to 0} {f(x)} + \lim_{x \to 0} {g(x)} = \lim_{x \to 0} \frac{x^3}{x} + \lim_{x \to 0} \frac{x^2}{x} = 0$$

f(x) + g(x) = 0
f(0) + g(0) = undefined

Is that correct?

 

[Mark44]

Almost. f(x) + g(x) = x² + x, if x is not 0.

 

[PhysicsAnonnn]

Almost. f(x) + g(x) = x² + x, if x is not 0.

How would you write that as a complete answer?


I would also like to know if:

$$\infty + a finite number = \infty$$

undefined + defined = undefined

 

[Char. Limit]

How would you write that as a complete answer?


I would also like to know if:

$$\infty + a = \infty$$

undefined + defined = undefined

Yes.

 

[PhysicsAnonnn]

Thank you for your help, Char. Limit, HallsofIvy, and Mark44.