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

[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

 

[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:

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.

 

[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.

 

[PeroK]

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.

 

[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.)

 

[PeroK]

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?

 

[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.

 

[PeroK]

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.

 

[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.