# Proof

(related to Theorem: Squeezing Theorem for Functions)

• Let $D\subseteq\mathbb R$ be a subset of real numbers.
• Let $f,g,h:D\to\mathbb R$ be functions with $g(x)\le f(x)\le h(x)$ for all $x\in D$.
• Let $g,h$ have a limit $L$ at $x=a\in D,$ i.e. $\lim_{x\to a} g(x)=\lim_{x\to a}h(x)=L.$
• Let $\epsilon > 0$ be given.
• By the definition of limit, there is a $\delta_1 > 0$ such that for all $x\in D$ with $0 < |x-a| < \delta_1,$ it follows $|g(x)-L| < \epsilon.$
• Similarly, there is a $\delta_2 > 0$ such that for all $x\in D$ with $0 < |x-a| < \delta_2,$ it follows $|h(x)-L| < \epsilon.$
• Take the minimum $\delta:=\min(\delta_1,\delta_2).$
• For $x\in D$ with $0 < |x-a| < \delta,$ it follows $|g(x)-L| < \epsilon$ and $|h(x)-L| < \epsilon.$
• Thus, for those $x,$ we have that $-\epsilon < g(x)-L\le f(x)-L\le h(x)-L < \epsilon.$
• For those $x,$ it follows that $|f(x)-L| < \epsilon.$
• By the definition of limit, $\lim_{x\to a} f(x)=L.$

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### References

#### Bibliography

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