# Proof

### Hypothesis

• Let $f,g$ have the limits $\lim_{x\to a}f(x)=L$ and $\lim_{x\to a}g(x)=H.$
• Given $\epsilon > 0.$

### Implications

• By the definition of limit:
• there is a $\delta_1 > 0$ such that if $x\in D$ and $0 < |x-a| < \delta_1,$ then $|f(x)-L|\le \frac \epsilon{2(|H|+1)},$ and
• there is a $\delta_2 > 0$ such that if $x\in D$ and $0 < |x-a| < \delta_2,$ then $|g(x)-H|\le \frac \epsilon{2(|L|+1)}.$
• Moreover, there is a $\delta_3 > 0$ sucht that if $x\in D$ and $0 < |x-a| < \delta_3,$ then $|f(x)-L| < 1.$ Thus, by the triangle inequality, $|f(x)|=|L-(f(x)-L)|\le |L|+|f(x)-L| < |L| + 1.$
• Get the minimum $\delta=\min(\delta_1,\delta_2,\delta_3).$
• Note that $\delta > 0.$
• Since $f\cdot g:D\to\mathbb R,$ we have that if $x\in D$ with $0 < |x-a| < \delta,$ then by the triangle inequality. $$\begin{array}{rcl}|(f(x)\cdot g(x))-(L\cdot H)|&=&|f(x)g(x)-f(x)H+f(x)H-LH)|\\ &=&|(f(x)(g(x)-H)+H(f(x)-L)|\\ &\le&|(f(x)(g(x)-H)|+|H(f(x)-L)|\\ &=&|(f(x)|\cdot |(g(x)-H)|+|H|\cdot |(f(x)-L)|\\ &=&||L|+1|\cdot\frac{\epsilon}{2(|L|+1)}+|H|\cdot\frac{\epsilon}{2(|H|+1)}\\ & < & \frac \epsilon2 + \frac \epsilon2\\ &=&\epsilon\end{array}.$$

### Conclusion

• By the definition of limit, this shows that the limit of the product of both functions is given by $\lim_{x\to a}(f(x)\cdot g(x))=L\cdot H.$

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### References

#### Bibliography

1. Kane, Jonathan: "Writing Proofs in Analysis", Springer, 2016