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

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1. Jan 11, 2017

issacnewton

1. The problem statement, all variables and given/known data
If $\lim_{x \rightarrow a} f(x) = \infty$ and $\lim_{x \rightarrow a} g(x) = c$, and if $c>0$ then prove that
$\lim_{x \rightarrow a} \left[ f(x)g(x)\right] = \infty~~ \text{if c > 0}$

2. Relevant equations
Epsilon Delta definition of the limit

3. 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

2. Jan 11, 2017

PeroK

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:

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.

3. Jan 11, 2017

issacnewton

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.

4. Jan 11, 2017

PeroK

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

5. Jan 11, 2017

issacnewton

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.)

6. Jan 11, 2017

PeroK

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?

7. Jan 11, 2017

issacnewton

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.

8. Jan 11, 2017

PeroK

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

9. Jan 11, 2017

issacnewton

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.