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Proof

(related to Proposition: Arithmetic of Functions with Limits - Sums)

Context

• Let $D\subseteq R$ be a subset of real numbers with an accumulation point $a.$
• Let $f,g:D\to\mathbb R$ be functions.

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 $|f(x)-L|\le \frac \epsilon2,$ and
  • there is a $\delta_2 > 0$ such that if $x\in D$ and $0 < |x-a| < \delta_2,$ then $|g(x)-H|\le \frac \epsilon2.$
• Get the minimum $\delta=\min(\delta_1,\delta_2).$
• Note that $\delta > 0.$
• Since $f+g:D\to\mathbb R,$ we have that if $x\in D$ with $0 < |x-a| < \delta,$ then by the triangle inequality $$|(f(x)+g(x))-(L+H)|=|(f(x)-L)+(g(x)-H)|\le |f(x)-L|+|g(x)-H| < \frac \epsilon2 + \frac \epsilon2=\epsilon.$$

Conclusion

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

Bibliography

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