# If f(x) -> oo and g(x) -> c, prove that f(x)+g(x) -> oo

• issacnewton
In summary, this conversation discusses the proof of a limit theorem using the epsilon-delta definition. It states that if two limits are given, one approaching infinity and one approaching a constant, the sum of the two limits will also approach infinity. The proof involves considering two cases and using the properties of limits to arrive at the conclusion. The proof is valid for all real numbers and is summarized in a formal statement at the end.
issacnewton

## Homework Statement

If ##\lim_{x \rightarrow a} f(x) = \infty ## and ## \lim_{x \rightarrow a} g(x) = c ##, then prove that
$$\lim_{x \rightarrow a} f(x) +g(x) = \infty$$

## Homework Equations

Epsilon Delta definition of limit

## The Attempt at a Solution

I will take two cases here.
Case 1. ## c \geqslant 0##
Let ##M>0## be arbitrary. Then ## 2M > 0##. Since we have ##\lim_{x \rightarrow a} f(x) = \infty ## , ##\exists~\delta_1 >0 ## such that
$$\forall~x \in D(f) \left[ 0 < |x-a| < \delta_1 \rightarrow f(x) > 2M \right] \cdots\cdots (1)$$
Where, ##D(f)## is the domain of function ##f##. We also have ##\frac M 2 > 0##. Since ## \lim_{x \rightarrow a} g(x) = c ##, ##\exists~\delta_2 >0 ## such that
$$\forall~x \in D(g) \left[ 0 < |x-a| < \delta_2 \rightarrow |g(x)-c| < \frac M 2 \right] \cdots\cdots (2)$$
where ##D(g)## is the domain of function ##g##. Now let ##\delta = \text{min}(\delta_1, \delta_2)##. Let ##x_1 \in D(f+g)## be arbitrary. And also suppose that ## 0 < |x-a| < \delta ##. Since ##x_1 \in D(f+g)##, we have that ## x_1 \in D(f) ## and ## x_1 \in D(g) ##. Also since ## 0 < |x-a| < \delta ##, from the definition of
##\delta##, it implies that ## 0 < |x-a| < \delta_1## and ## 0 < |x-a| < \delta_2 ##. So we can conclude from the equation 1 and 2 that, ## f(x_1) > 2M## and ## |g(x_1) -c| < \frac M 2##. It follows that $$c - \frac M 2 < g(x_1) < c + \frac M 2$$ $$\therefore f(x_1) + c - \frac M 2 <f(x_1)+ g(x_1) < f(x_1) + c + \frac M 2 \cdots\cdots(3)$$ But we have ##f(x_1) > 2M##, so ##f(x_1) - \frac M 2 > \frac{3M}{2}##. This leads to ## f(x_1) - \frac M 2 + c > \frac{3M}{2} + c##. Since ## c \geqslant 0##, we get ## \frac{3M}{2} + c \geqslant \frac{3M}{2} > M##. We conclude that ## f(x_1) - \frac M 2 + c > M##. Using this in equation ##(3)##, we have $$M < f(x_1) + c - \frac M 2 < f(x_1)+ g(x_1)$$ So we conclude that ## f(x_1)+ g(x_1) > M ##. Since ##M## and ##x_1## are arbitrary, we conclude that $$\lim_{x \rightarrow a} f(x) +g(x) = \infty$$ Now we proceed to the case 2.
Case 2. ## c<0##
Again let ##M > 0## be arbitrary. So we have ##-2c >0## and ##2M >0##. So ##2M - 2c >0##. Since we have ##\lim_{x \rightarrow a} f(x) = \infty ## , ##\exists~\delta_1 >0 ## such that $$\forall~x \in D(f) \left[ 0 < |x-a| < \delta_1 \rightarrow f(x) > (2M-2c) \right] \cdots\cdots (4)$$ Where, ##D(f)## is the domain of function ##f##. Since ##-c>0##, and since ## \lim_{x \rightarrow a} g(x) = c ##, ##\exists~\delta_2 >0 ## such that $$\forall~x \in D(g) \left[ 0 < |x-a| < \delta_2 \rightarrow |g(x)-c| < -c \right] \cdots\cdots (5)$$ where ##D(g)## is the domain of function ##g##. Now let ##\delta = \text{min}(\delta_1, \delta_2)##. Let ##x_1 \in D(f+g)## be arbitrary. And also suppose that ## 0 < |x-a| < \delta ##. Since ##x_1 \in D(f+g)##, we have that ## x_1 \in D(f) ## and ## x_1 \in D(g) ##. Also since ## 0 < |x-a| < \delta ##, from the definition of ##\delta##, it implies that ## 0 < |x-a| < \delta_1## and ## 0 < |x-a| < \delta_2 ##. So we can conclude from the equation 4 and 5 that, ## f(x_1) > 2M-2c ## and ## |g(x_1) -c| < -c ##. So we get ## c < g(x_1) - c < -c##. Which is ## 2c < g(x_1) < 0 ##. Therefore, ## f(x_1) + 2c < f(x_1) + g(x_1)##. Since## f(x_1) > 2M-2c ##, we have $$M< 2M < f(x_1)+2c < f(x_1) + g(x_1)$$ From here we conclude that ## f(x_1) + g(x_1) > M##. Since ##M## and ##x_1## are arbitrary, we conclude that $$\lim_{x \rightarrow a} f(x) +g(x) = \infty$$ So we have proven this for both cases and hence ##\forall ~c \in \mathbb{R}##, we can say that $$\lim_{x \rightarrow a} f(x) +g(x) = \infty$$ Can you check if this is OK!
Thanks

Seems correct to me but I don't see the reason you introduce ##x_1##, you should just say that for every x such that ##0<|x-a|<\delta## and proceed with x instead of x1.

Well Delta, Since we have a universal quantifier in the limit definition, I choose an arbitrary ##x_1## and then proceed to prove the statement inside that quantifier. After we prov the statement , then we can say that the statement is true for arbitrary ##x_1##, it is true for every x in the domain

Ok then I see a little mistake you say ##x_1## arbitrary but it isn't completely arbitrary it is such that ##0<|x_1-a|<\delta## so that (1) (2) or (4) and (5) are true for x1. You are saying this but you using x instead of x1.

Delta , I think the original goal here is to prove that $$\forall ~M >0 \exists~ \delta >0~ \forall ~x \in D(f+g)\left[ 0<|x-a| < \delta \rightarrow f(x)+g(x) > M \right]$$ So initially, I let ##M > 0## be arbitrary. Then with the choice of ##\delta## I have done, my goal becomes, $$\forall ~x \in D(f+g)\left[ 0<|x-a| < \delta \rightarrow f(x)+g(x) > M \right]$$ Since there is a universal quantifier at the beginiing, I let ## x \in D(f+g)## as arbitrary. So now the goal becomes $$0<|x-a| < \delta \rightarrow f(x)+g(x) > M$$ Since this is an implication, I assume the antecedent, ## 0<|x-a| < \delta## and then proceed to prove the consequent, which is ## f(x)+g(x) > M##.

Fine it is just that you assume the antecedent for ##x## while you prove the consequent for ##x_1##.

Actually now I see your point. I have used ##x_1## when I chose an arbitrary member in the domain , but I used ##x## in the antecedent. I should have changed that to ##x_1##. So with that error removed proof seems ok !

Delta2

## 1. How can you prove that f(x)+g(x) approaches infinity?

In order to prove that f(x)+g(x) approaches infinity, we can use the definition of limit. We know that when f(x) approaches infinity, it means that for any large positive number M, there exists a corresponding x-value where f(x) is greater than M. Similarly, when g(x) approaches a constant c, it means that for any positive number epsilon, there exists a corresponding x-value where g(x) is within epsilon distance of c. Combining these two definitions, we can show that for any large positive number N, there exists a corresponding x-value where f(x)+g(x) is greater than N+c, thus approaching infinity.

## 2. Is it necessary for both f(x) and g(x) to approach infinity and a constant, respectively, in order for f(x)+g(x) to approach infinity?

Yes, in order for f(x)+g(x) to approach infinity, both f(x) and g(x) must approach infinity and a constant, respectively. This is because if only one of them approaches infinity, the sum may still be bounded and not approach infinity. For example, if f(x) approaches infinity and g(x) approaches a constant c, the sum f(x)+g(x) will approach infinity. However, if both f(x) and g(x) approach infinity, the sum will also approach infinity.

## 3. Can we use the Squeeze Theorem to prove that f(x)+g(x) approaches infinity?

No, the Squeeze Theorem cannot be used to prove that f(x)+g(x) approaches infinity. The Squeeze Theorem is used to prove that a function approaches a certain limit by showing that it is always between two other functions that approach the same limit. In this case, we are trying to prove that the sum of two functions approaches infinity, not a specific limit.

## 4. What if f(x) approaches infinity and g(x) approaches negative infinity?

In this case, the sum f(x)+g(x) may or may not approach infinity, depending on the rate at which each function approaches their respective infinities. If one function approaches infinity at a faster rate than the other approaches negative infinity, the sum will approach infinity. However, if one function approaches negative infinity faster than the other approaches infinity, the sum will approach negative infinity instead.

## 5. Is it possible for f(x)+g(x) to approach infinity if f(x) and g(x) approach negative infinity?

Yes, it is possible for f(x)+g(x) to approach infinity even if f(x) and g(x) both approach negative infinity. This can happen if the sum of the two functions approaches 0 at a very slow rate. For example, if f(x) approaches negative infinity at a slower rate than g(x) approaches negative infinity, the sum f(x)+g(x) will approach infinity.

Replies
8
Views
2K
Replies
6
Views
2K
Replies
9
Views
980
Replies
1
Views
920
Replies
2
Views
2K
Replies
2
Views
1K
Replies
32
Views
2K
Replies
6
Views
782
Replies
2
Views
2K
Replies
5
Views
1K