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## Homework Statement

Prove [itex]f(x)=\sqrt{x^{2}+1}[/itex] is uniformly continuous on the real line.

## Homework Equations

Lipschitz Condition: If there is a constant [itex]M[/itex] such that [itex]|f(p) - f(q)| \leq M |p-q|[/itex] for all [itex]p,q \in D[/itex], then [itex]f[/itex] obeys the Lipschitz condition.

Mean Value Theorem: Let [itex]f[/itex] be continuous on [itex][a,b][/itex] and let [itex]f'(x)[/itex] exist for all [itex]x \in (a,b)[/itex]. Then at least one point [itex]x_{o}[/itex] exists in [itex](a,b)[/itex] such that [itex]f(b) - f(a) = f'(x_{o})(b-a)[/itex].

[itex]f'(x) = \frac{x}{\sqrt{x^{2} + 1}}[/itex] is bounded below by [itex]-1[/itex] and bounded above by [itex]1[/itex].

## The Attempt at a Solution

Proof:

[itex]f(x)=\sqrt{x^{2}+1}[/itex] is continuous on [itex][-N,N][/itex] and differentiable on [itex](-N,N)[/itex] for all [itex]N[/itex]. By the Mean Value Theorem, there exists an [itex]x_{o} \in (-N,N)[/itex] such that [itex]|f(x)-f(y)| \leq |f'(x_{o})||x-y|[/itex] for all [itex]x,y \in (-N,N)[/itex].

Since the derivative [itex]f'(x_{o})[/itex] is bounded above by [itex]1[/itex] and below by [itex]-1[/itex] as [itex]x[/itex] tends to infinity, [itex]f'(x)[/itex] obeys [itex]0 < |f'(x)| < 1[/itex] for all [itex]x \in (-N,N)[/itex]. Thus [itex]0 < |f(x)-f(y)| \leq 1 \cdot |x-y|[/itex] for all [itex]x,y \in (-N,N)[/itex].

Since these results hold for a general [itex]N[/itex], we may choose a larger [itex]N[/itex] and the results will still hold for the larger [itex]N[/itex], implying the results will hold for the real line as a whole. Letting [itex]\delta < \epsilon[/itex] thus shows that [itex]f(x)[/itex] is uniformly continuous on the real line.

QED

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