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# proof of limit rule of product

Let $f$ and $g$ be real 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.

Keywords:

limit rule of product

Related:

ProductOfFunctions, TriangleInequality, ProductAndQuotientOfFunctionsSum

Type of Math Object:

Proof

Major Section:

Reference

Parent:

Groups audience:

## Mathematics Subject Classification

30A99*no label found*26A06

*no label found*

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