Prove that if \begin{equation} \lim_{x\to\infty}\,f(x)=L \end{equation} and \begin{equation} \lim_{x\to\infty}\,g(x)=K \end{equation} then \begin{equation} \lim_{x\to\infty}\,f(x)\,g(x)=LK. \end{equation}

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Lesson Parent: 
Function Limits IX: Horizontal Asymptotes And The Limit Of $f(x)$ As $x$ Tends To Infinity
Problem: 

Prove that if
\begin{equation}
\lim_{x\to\infty}\,f(x)=L
\end{equation}
and
\begin{equation}
\lim_{x\to\infty}\,g(x)=K
\end{equation}
then
\begin{equation}
\lim_{x\to\infty}\,f(x)\,g(x)=LK.
\end{equation}

SAMPLE SOLUTION

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Solution: 

First, from our rules of inequalities we know that

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Because

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we know that there exists a $Y=Y_1$ such that <Sign in to see all the formulas> and thus <Sign in to see all the formulas>. Furthermore, we also know that there exists a $Y=Y_2$ such that <Sign in to see all the formulas>.

Because

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we know that there exists a $Y=Y_3$ such that <Sign in to see all the formulas>.

So now we choose the largest of the three $Y$'s (<Sign in to see all the formulas>) so that

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Finally, we have shown that given any <Sign in to see all the formulas> there exits a <Sign in to see all the formulas> such that <Sign in to see all the formulas>. Therefore,

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There is one important subtlety that we want to point out. We first pointed out that

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The important thing to note is that we used <Sign in to see all the formulas> and not <Sign in to see all the formulas>. If we had used <Sign in to see all the formulas> then we would have found a $Y_2$ such that <Sign in to see all the formulas>. On the surface this seems fine. However, what if $K=0$? Then <Sign in to see all the formulas> would have been infinite and we could not say <Sign in to see all the formulas> is less than any specific number and therefore not been able to say that <Sign in to see all the formulas> would have been less than any specific number. To keep the proof usable for any value of $K$ (even when $K$ is zero) then we use <Sign in to see all the formulas> which means <Sign in to see all the formulas> would be bounded even for $K=0$.