# A question about limits at infinity

I got this question:

Let $f, g$ be functions defined on the interval $[0,\infty)$ and let $L \in \mathbb{R}$ be a real number such that $$\lim_{x \to \infty}\left(f(x)\cdot g(x)\right)= L.$$

(1) If $\lim_{x \to \infty}f(x) = m$ where $0 \neq m \in \mathbb{R}$ is a nonzero real number, Must it be the case that $\lim_{x \to \infty}g(x)$ exist?

(2) If $\lim_{x \to \infty}f(x) = \infty$ Must it be the case that $\lim_{x \to \infty}g(x)$ exist?

I having hard time trying to prove it and to find a counter example. Thanks.

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## 2 Answers

We know that if $\lim\limits_{x\rightarrow a}{y(x)}=a$ and $\lim\limits_{x\rightarrow a}{w(x)}=b$ then $\lim\limits_{x\rightarrow a}{\left(y(x)\times w(x)\right)}=ab$

Now $\lim\limits_{x\rightarrow \infty}{\dfrac1{f(x)}}=m$ (as $m$ is non-zero)

Consider functions $y(x)=\dfrac{1}{f(x)}$ and $w(x)=f(x)\times g(x)$. Now as both the limits of $y(x)$ and $w(x)$ exists. If you consider the limit of $y(x)\times w(x)\color{grey}{=g(x)}$, you can prove that $g(x)$ approaches the limit $\dfrac{L}{m}$.

About the second question:

• As you might have noticed, before defining the limit, we are imposing a condition that $L$ is real I want to avoid things like $\lim{f(x)}=\infty$ and claiming that $\lim f(x)$ exist .We write $f(x)\rightarrow +\infty$ as $x\rightarrow a$ instead of the $\lim$ equals $\infty$). Now this becomes tricky! Although it looks like the previous proof works for this one too, it is not so. It cannot be used as the product of the limits equals the limit of product is applicable only if the limit of both the functions 'exist', the existence of limit of $f(x)$ is a problem here, so we have to resolve to more elementary technique.

We say that a function $f(x)$ approaches $\infty$ if for every $M\in \mathbb{R}$ we can find an $N\in \mathbb{R}$ such that $$x\ge N \implies f(x)\ge M.$$

Now as $f(x)g(x)$ approaches $L$ we can say that for every $\varepsilon^{'} =\varepsilon \cdot M -L$ there exists an $N_1$ such that $x>N_1 \implies -\varepsilon<\frac{L-\varepsilon^{'}}{f(x)}<g(x)<\frac{L+\varepsilon^{'}}{f(x)} <\varepsilon \tag{1}$

And thus $g(x)$ approaches $0$

(For proving (1) you will have to assume that $L\ge0$, for the case $L<0$ you can consider $\varepsilon^{'}=\varepsilon \cdot M +L$ and arrive at the same inequality. Also division by $f(x)$ is valid as we can have an $N_2$ such that for all $x>N_2$, $f(x)$ is greater than $0$)

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Very exhaustive. – Martín-Blas Pérez Pinilla Mar 27 '14 at 12:49
Thank you very much. – MathNerd Mar 27 '14 at 14:09
Thank you, now I finished proving the whole thing and it took me 3 full pages, much thanks. – MathNerd Mar 27 '14 at 16:31

$$\lim_{x\to\infty}g(x)=\lim_{x\to\infty}{f(x)g(x)\over f(x)}= {\lim_{x\to\infty}f(x)g(x)\over\lim_{x\to\infty}f(x)}={L\over m}.$$ (why we can divide by $f(x)$?)

If (2), then $$\lim_{x\to\infty}g(x)=\lim_{x\to\infty}{f(x)g(x)\over f(x)}=0$$ because (bounded thing)/(thing$\to\infty$)$\to 0$.

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Thanks, I tried to prove it using epsilon delta but forgot to use arithmetic of limits. – MathNerd Mar 27 '14 at 11:55
@Martin I disagree that the same can be used for the second question. Please read the answer I have posted. – GNU Emacs Mar 27 '14 at 12:46
(Bounded thing)/(thing $\to\infty$)$\to 0$. – Martín-Blas Pérez Pinilla Mar 27 '14 at 12:48
@Martín-BlasPérezPinilla: You start with saying that $\lim_{x\to\infty}{f(x)g(x)\over f(x)}= {\lim_{x\to\infty}f(x)g(x)\over\lim_{x\to\infty}f(x)}$, that cannot be considered to be holding in the second case. – GNU Emacs Mar 27 '14 at 12:50
I've edited my answer. – Martín-Blas Pérez Pinilla Mar 27 '14 at 12:54