If f(x) -> oo and g(x) -> c, then f(x)g(x) -> -oo if c < 0

• issacnewton
In summary: Vickson. I will keep that in mind.In summary, the conversation discusses a proof involving epsilon-delta definition and limits of functions. The result is that if the limit of one function approaches infinity and the limit of another function approaches a negative number, then the limit of their product is negative infinity. One person suggests that this result can be deduced from a previous result, rather than repeating the same steps. The other person agrees and suggests focusing on developing insight rather than just honing skills.
issacnewton

Homework Statement

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

Homework Equations

Epsilon Delta definition

The Attempt at a Solution

Let ##N<0## be arbitrary. Since ##c<0##, we have ##\frac{2N} c >0##. Since ##\lim_{x \rightarrow a} f(x) =\infty##, ##\exists~ \delta_1>0## such that $$\forall ~ x\in D(f)\left[ 0<|x-a| <\delta_1 \rightarrow f(x) > \frac{2N} c \right]\cdots\cdots(1)$$ Also ##-\frac c 2 >0##, and since ## \lim_{x \rightarrow a} g(x) = c##, ##\exists ~ \delta_2 >0 ## such that $$\forall ~x \in D(g) \left[ 0<|x-a| <\delta_2 \rightarrow |g(x) - c| < -\frac c 2 \right]\cdots\cdots(2)$$ Here ##D(f)## and ##D(g)## are the domains of functions ##f## and ##g##. Let ##\delta = \text{min}(\delta_1, \delta_2)## and let ##x_1 \in D(fg)## be arbitrary, where ##D(fg)## is the domain of function ##fg##. So we have ##x_1 \in D(f)## and ##x_1 \in D(g)##. Also suppose ##0<|x_1-a| <\delta ##. It follows that ##0<|x_1-a| <\delta_1## and ## 0<|x_1-a| <\delta_2##. So now using equations ##(1)## and ##(2)##, we deduce that ##f(x_1) > \frac{2N} c ## and ##|g(x_1)-c| < -\frac c2## It follows that ##g(x_1) < \frac c 2##. Since ##f(x_1) > \frac{2N} c > 0##, we get ## f(x_1)g(x_1) <f(x_1) \frac c 2##. But as ##f(x_1) > \frac{2N} c## , we have ##f(x_1)\frac c 2 < N##. It follows that ##f(x_1)g(x_1) < N##. As ##N<0## is arbitrary, and ##x_1 \in D(fg)## is arbitrary, from the definition it follows that $$\lim_{x \rightarrow a} \left[ f(x)g(x)\right] = -\infty ~\text{ if c < 0 }$$ Please check my proof. Thanx

Given your previous post, I would say there is no need to prove this explicitly, as it is just the same steps repeated. This results follows from the previous result. I think that is a better mathematical approach to deduce a similar result from a previous one than to go back to first principles and effectively repeat the same steps.

Perok, I could have done that. But just wanted to hone my ##\epsilon-\delta## skills in these kinds of proofs.

IssacNewton said:
Perok, I could have done that. But just wanted to hone my ##\epsilon-\delta## skills in these kinds of proofs.

Okay, but perhaps it's time to move on to more challenging problems.

Slowly I am moving there

IssacNewton said:
Perok, I could have done that. But just wanted to hone my ##\epsilon-\delta## skills in these kinds of proofs.

But you are doing that at the expense of insight. You have proven the result already if ##g(x) \to c > 0##, so to know the result for ##g(x) \to c < 0## just apply the previous result to ##h(x) = -g(x)##, in which ##h(x) \to -c > 0##. That IS a proof!

When I mention "insight" I mean that someone wanting to become adept at Mathematics should try very hard to recognize similar cases already solved, and just apply previous results whenever possible. Repeating over and over the same type of argument does not enhance one's understanding, and might even be regarded as working against true understanding.

Last edited:
fresh_42 and PeroK
Thanks Ray

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

The notation "f(x) -> oo" means that the function f(x) approaches infinity as x approaches a certain value. In other words, the value of f(x) gets larger and larger without bound as x gets closer and closer to the specified value.

2. How does this statement apply to the limit of f(x)g(x)?

If both f(x) and g(x) approach infinity and a constant value c, respectively, then the limit of their product, f(x)g(x), will approach negative infinity if c is less than 0. This is because as f(x) gets larger and larger, g(x) gets closer and closer to a constant value, resulting in a product that gets increasingly negative.

3. Can f(x) and g(x) approach different values?

Yes, f(x) and g(x) can approach different values as long as f(x) approaches infinity and g(x) approaches a constant value c. The statement holds true as long as c is less than 0.

4. What happens if c is greater than 0?

If c is greater than 0, then the limit of f(x)g(x) will approach positive infinity. This is because as f(x) gets larger and larger, g(x) also gets larger and larger, resulting in a product that approaches infinity.

5. Is there a specific value that c must be for this statement to be true?

No, there is no specific value that c must be for this statement to be true. As long as c is less than 0, the limit of f(x)g(x) will approach negative infinity. However, if c is greater than or equal to 0, the limit will approach positive infinity.

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