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

[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

 

[PeroK]

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.

 

[issacnewton]

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

 

[PeroK]

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.

 

[issacnewton]

Slowly I am moving there

 

[Ray Vickson]

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.

 

[issacnewton]

Thanks Ray