# proof of limit rule of product

Let $f$ and $g$ be real (http://planetmath.org/RealFunction) or complex functions having the limits

 $\lim_{x\to x_{0}}f(x)=F\quad\mbox{and}\quad\lim_{x\to x_{0}}g(x)=G.$

Then also the limit $\displaystyle\lim_{x\to x_{0}}f(x)g(x)$ exists and equals $FG$.

Proof.  Let $\varepsilon$ be any positive number.  The assumptions imply the existence of the positive numbers $\delta_{1},\,\delta_{2},\,\delta_{3}$ such that

 $\displaystyle|f(x)-F|<\frac{\varepsilon}{2(1+|G|)}\;\;\mbox{when}\;\;0<|x-x_{0% }|<\delta_{1}$ (1)
 $\displaystyle|g(x)-G|<\frac{\varepsilon}{2(1+|F|)}\;\;\mbox{when}\;\;0<|x-x_{0% }|<\delta_{2},$ (2)
 $\displaystyle|g(x)-G|<1\;\;\mbox{when}\;\;0<|x-x_{0}|<\delta_{3}.$ (3)

According to the condition (3) we see that

 $|g(x)|=|g(x)\!-\!G\!+\!G|\leqq|g(x)\!-\!G|+|G|<1\!+\!|G|\;\;\mbox{when}\;\;0<|% x-x_{0}|<\delta_{3}.$

Supposing then that  $0<|x-x_{0}|<\min\{\delta_{1},\,\delta_{2},\,\delta_{3}\}$  and using (1) and (2) we obtain

 $\displaystyle|f(x)g(x)-FG|$ $\displaystyle=|f(x)g(x)-Fg(x)+Fg(x)-FG|$ $\displaystyle\leqq|f(x)g(x)\!-\!Fg(x)|+|Fg(x)\!-\!FG|$ $\displaystyle=|g(x)|\cdot|f(x)\!-\!F|+|F|\cdot|g(x)\!-\!G|$ $\displaystyle<(1\!+\!|G|)\frac{\varepsilon}{2(1\!+\!|G|)}+(1\!+\!|F|)\frac{% \varepsilon}{2(1\!+\!|F|)}$ $\displaystyle=\varepsilon$

This settles the proof.

Title proof of limit rule of product ProofOfLimitRuleOfProduct 2013-03-22 17:52:22 2013-03-22 17:52:22 pahio (2872) pahio (2872) 6 pahio (2872) Proof msc 30A99 msc 26A06 ProductOfFunctions TriangleInequality ProductAndQuotientOfFunctionsSum