Proof of addition of limits when the output value is infinite

In summary: Let's say you have an equation for f(x) = g(x), and you want to find the limit as x approaches a. What you need to do is solve the equation for g(x) as x approaches a, and then use the limit to find the limit of f(x) as x approaches a.It shouldn't be hard to prove.
  • #1
swampwiz
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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?
 
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  • #2
swampwiz said:
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?
 
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  • #3
PeroK said:
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.
 
  • #4
swampwiz said:
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.
 
  • #5
Mark44 said:
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
 
  • #6
swampwiz said:
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:
$$\text {If} \ \lim_{x \rightarrow a} f(x) = +\infty \ \text{and} \ \lim_{x \rightarrow a} g(x) = +\infty, $$
$$\ \text{then} \ \lim_{x \rightarrow a} (f(x)+g(x)) = +\infty$$
 
  • #7
swampwiz said:
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.
 
  • #8
PeroK said:
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.
 
  • #9
swampwiz said:
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.
 

FAQ: Proof of addition of limits when the output value is infinite

1. What is the proof of addition of limits when the output value is infinite?

The proof of addition of limits when the output value is infinite is a mathematical concept that shows how to find the limit of the sum of two functions when one or both of the functions have an infinite output value.

2. Why is it important to understand the proof of addition of limits when the output value is infinite?

Understanding the proof of addition of limits when the output value is infinite is important because it allows us to accurately determine the limit of a sum of functions, even when one or both of the functions have an infinite output value. This is crucial in many mathematical and scientific applications.

3. How is the proof of addition of limits when the output value is infinite different from the proof of addition of limits when the output value is finite?

The proof of addition of limits when the output value is infinite is different from the proof of addition of limits when the output value is finite because it involves using special techniques, such as L'Hôpital's rule, to handle the infinite output values. In contrast, the proof for finite output values can be done using basic algebraic manipulation.

4. Can you provide an example of the proof of addition of limits when the output value is infinite?

Sure, let's say we have the functions f(x) = 1/x and g(x) = 1/x^2. Both of these functions have an infinite output value as x approaches 0. Using the proof of addition of limits, we can show that the limit of f(x) + g(x) as x approaches 0 is also infinite.

5. Are there any limitations to the proof of addition of limits when the output value is infinite?

Yes, there are limitations to the proof of addition of limits when the output value is infinite. This proof only applies to functions that have an infinite output value at the same point. It cannot be used for functions with different points of infinity or for functions with different types of infinity, such as positive and negative infinity.

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