Properties of Limit at Infinity of Real Function/Product Rule

Theorem

Let $a \in \R$.

Let $f, g : \hointr a \infty \to \R$ be real functions such that:

$\ds \lim_{x \mathop \to \infty} \map f x = L_1$

and:

$\ds \lim_{x \mathop \to \infty} \map g x = L_2$

where $\ds \lim_{x \mathop \to \infty}$ denotes the limit at $+\infty$.

Then:

$\ds \lim_{x \mathop \to \infty} \paren {\map f x \map g x} = L_1 L_2$

Proof

Write:

\(\ds \size {\map f x \map g x - L_1 L_2}\)

\(=\)

\(\ds \size {\map f x \map g x - L_1 L_2 + L_2 \map f x - L_2 \map f x}\)

\(\ds \)

\(=\)

\(\ds \size {\map f x \paren {\map g x - L_2} + L_2 \paren {\map f x - L_1} }\)

\(\ds \)

\(\le\)

\(\ds \size {\map f x} \size {\map g x - L_2} + \size {L_2} \size {\map f x - L_1}\)

Since:

$\ds \lim_{x \mathop \to \infty} \map f x = L_1$

there exists a real number $M_1 \ge 0$ such that:

$\size {\map f x - {L_1} } < 1$

for $x \ge M_1$.

By the Reverse Triangle Inequality, we then have:

$\size {\size {\map f x} - \size {L_1} } < 1$

so that:

$\size {L_1} - 1 < \size {\map f x} < \size {L_1} + 1$

So we have, for $x \ge M_1$:

$\size {\map f x} \size {\map g x - L_2} + \size {L_2} \size {\map f x - L_1} < \paren {\size {L_1} + 1} \size {\map g x - L_2} + \size {L_2} \size {\map f x - L_1}$

Since:

$\ds \lim_{x \mathop \to \infty} \map g x = L_2$

given $\epsilon > 0$ we can select a real number $M_2 \ge 0$ such that:

$\ds \size {\map g x - L_2} < \frac \epsilon {2 \paren {\size {L_1} + 1} }$ for $x \ge M_2$.

If $L_2 = 0$, we are done, since if $x \ge \max \set {M_1, M_2}$, we have:

\(\ds \size {\map f x \map g x}\)

\(<\)

\(\ds \paren {\size {L_1} + 1} \size {\map g x - L_2}\)

\(\ds \)

\(<\)

\(\ds \paren {\size {L_1} + 1} \paren {\frac \epsilon {2 \paren {\size {L_1} + 1} } }\)

\(\ds \)

\(=\)

\(\ds \frac \epsilon 2\)

Then since $\epsilon$ was arbitrary, we then have:

$\ds \lim_{x \mathop \to \infty} \paren {\map f x \map g x} = 0$

Now suppose that $L_2 \ne 0$.

Since:

$\ds \lim_{x \mathop \to \infty} \map f x = L_1$

given $\epsilon > 0$ we can pick a real number $M_3 \ge 0$ such that:

$\ds \size {\map f x - L_1} < \frac \epsilon {2 \size {L_2} }$

Let:

$M = \max \set {M_1, M_2, M_3}$

Then, for $x \ge M$, we have:

\(\ds \paren {\size {L_1} + 1} \size {\map g x - L_2} + \size {L_2} \size {\map f x - L_1}\)

\(<\)

\(\ds \paren {\size {L_1} + 1} \paren {\frac \epsilon {2 \paren {\size {L_1} + 1} } } + \size {L_2} \paren {\frac \epsilon {2 \size {L_2} } }\)

\(\ds \)

\(=\)

\(\ds \frac \epsilon 2 + \frac \epsilon 2\)

\(\ds \)

\(=\)

\(\ds \epsilon\)

$\blacksquare$