Proof of addition of limits when the output value is infinite

[swampwiz]

I was looking at some websites that show the proof of addition of limits for a finite output value, but I don't see one for the case of infinite output value, which has a different condition that needs to be met - i.e., | f( x ) | > M instead of | f( x ) - L | < ε.

http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx

http://www.milefoot.com/math/calculus/limits/GenericLimitLawProofs04.htm

And I can't use the trick of letting M be ½ since the triangle inequality doesn't work in the proper direction.

Any idea on how this proof is done for an infinite value?

 

[PeroK]

I was looking at some websites that show the proof of addition of limits for a finite output value, but I don't see one for the case of infinite output value, which has a different condition that needs to be met - i.e., | f( x ) | > M instead of | f( x ) - L | < ε.

http://tutorial.math.lamar.edu/Classes/CalcI/LimitProofs.aspx

http://www.milefoot.com/math/calculus/limits/GenericLimitLawProofs04.htm

And I can't use the trick of letting M be ½ since the triangle inequality doesn't work in the proper direction.

Any idea on how this proof is done for an infinite value?

What exactly are you trying to prove?

 

[swampwiz]

What exactly are you trying to prove?

AIUI, the ε-δ comparison is not { | f( x ) - L | < ε }, but rather { | f( x ) | > M }, so the trick of making each term be { f( x ) > ½ M } doesn't work.

 

[Mark44]

AIUI, the ε-δ comparison is not { | f( x ) - L | < ε }, but rather { | f( x ) | > M }, so the trick of making each term be { f( x ) > ½ M } doesn't work.

What is AIUI? If it's an acronym, it's not one I've seen before.

All you need to do is to show that, given any (large and positive) M, then you can find a $\delta > 0$ such that f(x) > M.

Try it with f(x) = 1/x, and $\lim_{x \to 0^+} f(x)$. If it's not obvious, can you find a number $\delta$ so that f(x) > 1000? 10,000? 100,000? You'll need to find a different $\delta$ for each.

 

[swampwiz]

What is AIUI? If it's an acronym, it's not one I've seen before.

All you need to do is to show that, given any (large and positive) M, then you can find a $\delta > 0$ such that f(x) > M.

Try it with f(x) = 1/x, and $\lim_{x \to 0^+} f(x)$. If it's not obvious, can you find a number $\delta$ so that f(x) > 1000? 10,000? 100,000? You'll need to find a different $\delta$ for each.

I understand when it's a single function, but the case I was referring to was when it's the sum of a pair of functions.

AIUI = As I Understand It

 

[PeroK]

I understand when it's a single function, but the case I was referring to was when it's the sum of a pair of functions.

AIUI = As I Understand It

AIUI you want to prove:
$$\text {If} \ \lim_{x \rightarrow a} f(x) = +\infty \ \text{and} \ \lim_{x \rightarrow a} g(x) = +\infty, $$
$$\ \text{then} \ \lim_{x \rightarrow a} (f(x)+g(x)) = +\infty$$

 

[Mark44]

I understand when it's a single function, but the case I was referring to was when it's the sum of a pair of functions.

If the limit is as @PeroK shows, then let h(x) = f(x) + g(x), and follow what I described earlier.

 

[swampwiz]

AIUI you want to prove:
$$\text {If} \ \lim_{x \rightarrow a} f(x) = +\infty \ \text{and} \ \lim_{x \rightarrow a} g(x) = +\infty, $$
$$\ \text{then} \ \lim_{x \rightarrow a} (f(x)+g(x)) = +\infty$$

Yes, although for all cases involving an infinity. I think the website I had mentioned goes into this after dealing with non-infinity cases. I will review that; I think it might explain what I am looking for.

 

[PeroK]

Yes, although for all cases involving an infinity. I think the website I had mentioned goes into this after dealing with non-infinity cases. I will review that; I think it might explain what I am looking for.

It shouldn't be hard to prove.