# As f(x) -> oo and g(x) -> c, if c > 0 then f(x)g(x) -> oo

• issacnewton
In summary, the proof shows that if the limit of one function is infinity and the limit of another function is a positive constant, then the limit of their product is also infinity. This is proven by using the epsilon-delta definition of the limit and choosing appropriate values for delta to show that the product of the two functions is greater than any arbitrary value, thus making their limit infinity. The proof also includes some intermediate algebraic steps to help with clarity for beginners.
issacnewton

## Homework Statement

If ## \lim_{x \rightarrow a} f(x) = \infty## and ##\lim_{x \rightarrow a} g(x) = c ##, and if ##c>0## then prove that
##\lim_{x \rightarrow a} \left[ f(x)g(x)\right] = \infty~~ \text{if c > 0}##

## Homework Equations

Epsilon Delta definition of the limit

## The Attempt at a Solution

Let ##M>0## be arbitrary. Since ##c>0##, we have ##\frac{2M} c >0## and ##\frac c 2 > 0##. Since ##\lim_{x \rightarrow a} f(x) = \infty## , ##\exists ~ \delta_1 >0## such that $$\forall ~ x \in D(f)\left[ 0 <|x-a| < \delta_1 \rightarrow f(x) > \frac{2M} c \right]\cdots\cdots(1)$$ Similarly, since ##\lim_{x \rightarrow a} g(x) = c ## , ##\exists ~ \delta_2 >0 ## such that $$\forall ~ x \in D(g)\left[ 0 <|x-a| < \delta_2 \rightarrow |g(x)-c| < \frac c 2 \right] \cdots\cdots(2)$$ Here ##D(f)## and ##D(g)## are the domains of the functions ##f## and ##g##. Now let ##\delta = \text{min}(\delta_1, \delta_2) ## and let ##x_1 \in D(fg)## be arbitrary element. Then ##x_1 \in D(f)## and ##x_1 \in D(g)##. Also suppose ##0 <|x_1-a| < \delta ##. It follows from the definition of ##\delta## that ##0 <|x_1-a| < \delta_1 ## and ##0 <|x_1-a| < \delta_2 ##. So using the equations 1 and 2, it follows that ##f(x_1) > \frac{2M} c## and ##|g(x_1) - c| < \frac c 2##. So we have ##\frac c 2 < g(x_1) < \frac{3c} 2##. Since ##g(x_1) > \frac c 2 > 0##, we can deduce that ## f(x_1) g(x_1) > g(x_1) \frac{2M} c## But ##g(x_1) \frac{2M} c > \frac c 2 \bullet \frac{2M} c ##. So it follows that ##f(x_1) g(x_1) > \frac c 2 \bullet \frac{2M} c ##. Hence ## f(x_1) g(x_1) > M##. Since ##M>0## is arbitrary and ##x_1 \in D(fg)## is arbitrary, it follows that $$\lim_{x \rightarrow a} \left[ f(x)g(x)\right] = \infty~~ \text{if c > 0}$$ Does this look correct ?
Thanks

Delta2
Just a comment on style. It's very good in general. At this level, however, you can start to be more succinct with the basic algebra. For example:

IssacNewton said:
So using the equations 1 and 2, it follows that ##f(x_1) > \frac{2M} c## and ##|g(x_1) - c| < \frac c 2##. So we have ##\frac c 2 < g(x_1) < \frac{3c} 2##. Since ##g(x_1) > \frac c 2 > 0##, we can deduce that ## f(x_1) g(x_1) > g(x_1) \frac{2M} c## But ##g(x_1) \frac{2M} c > \frac c 2 \bullet \frac{2M} c ##. So it follows that ##f(x_1) g(x_1) > \frac c 2 \bullet \frac{2M} c ##. Hence ## f(x_1) g(x_1) > M##.

I would do something like:

Equations 1 and 2 imply that ##f(x_1) > \frac{2M} c## and ##|g(x_1) - c| < \frac c 2##, hence ##g(x_1) > \frac{c}{2}## and ##f(x_1)g(x_1) > M##.

That's all you need.

Thanks PeroK, I see your point. But there might be people visiting this thread who may want details. This is an introductory Calculus problem I am doing here (from Stewart's Calculus 8edition, ex. 1.7.44). So those kind of students may want more clarity in the proofs.

IssacNewton said:
Thanks PeroK, I see your point. But there might be people visiting this thread who may want details. This is an introductory Calculus problem I am doing here (from Stewart's Calculus 8edition, ex. 1.7.44). So those kind of students may want more clarity in the proofs.

Excessive minor details can obscure a proof rather than clarify it.

I think taking lot of logical jumps in a proof can make it difficult to read too. I am just making sure that even beginners visiting this site will be able to understand this proof. Of course I am assuming the knowledge of basic logic (quantifiers, implication etc.)

IssacNewton said:
I think taking lot of logical jumps in a proof can make it difficult to read too. I am just making sure that even beginners visiting this site will be able to understand this proof. Of course I am assuming the knowledge of basic logic (quantifiers, implication etc.)

Why are you posting these if they are not your homework? If you know better, why ask our advice? What "beginners" are looking at your homework to learn analysis?

I am doing these problems myself. I just wanted feedback if they are correct. But I am also aware that people will visit this thread in the future and so wanted to make my proof as clear as possible.

IssacNewton said:
I am doing these problems myself. I just wanted feedback if they are correct. But I am also aware that people will visit this thread in the future and so wanted to make my proof as clear as possible.

Okay, but, believe it or not, it is harder to read with all those intermediate algebraic steps cluttering it up.

If the proof has too many logical jumps, I find it difficult to read. So you are right that I am cluttering it up but there is more clarity and some people who will visit this thread, who are beginners in analysis, will appreciate it I guess.

## 1. What does the notation "f(x) -> oo" mean?

The notation "f(x) -> oo" means that as the independent variable x approaches infinity, the function f(x) also approaches infinity. This is also known as the limit of f(x) as x approaches infinity.

## 2. How does the notation "g(x) -> c" differ from "f(x) -> oo"?

The notation "g(x) -> c" means that as x approaches infinity, the function g(x) approaches a constant value c. This is also known as the limit of g(x) as x approaches infinity.

## 3. What does it mean if c is greater than 0 in the statement "if c > 0 then f(x)g(x) -> oo"?

If c is greater than 0, it means that the limit of g(x) as x approaches infinity is a positive value. This information is important because it affects the behavior of the product f(x)g(x) as x approaches infinity.

## 4. How does the behavior of f(x) and g(x) affect the limit of their product, f(x)g(x)?

As stated in the previous question, the limit of f(x)g(x) is affected by the behavior of f(x) and g(x) as x approaches infinity. If both functions approach infinity (f(x) -> oo and g(x) -> oo), then their product will also approach infinity. However, if one function approaches a constant value while the other approaches infinity (f(x) -> oo and g(x) -> c), then their product will approach either a finite value or infinity depending on the value of c.

## 5. Can the statement "f(x)g(x) -> oo" be true if c is less than or equal to 0?

No, the statement "f(x)g(x) -> oo" only holds true when c is greater than 0. This is because if c is less than or equal to 0, then the limit of g(x) as x approaches infinity is either 0 or a negative value. In this case, the product f(x)g(x) would approach either a finite value or negative infinity, not positive infinity as stated in the original statement.

Replies
6
Views
2K
Replies
6
Views
2K
Replies
9
Views
973
Replies
1
Views
915
Replies
2
Views
1K
Replies
13
Views
819
Replies
17
Views
1K
Replies
8
Views
2K
Replies
1
Views
548
Replies
6
Views
759