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Question: Prove that $x+\frac x{x+1}$ is uniformly continuous in $(1,\infty)$:

My solution: We need to find a $\delta$ that satisfies the criterion for uniform continuity.

  • Let $\varepsilon >0 , x,y>1 $
  • We pick $\delta = \frac 4{5\varepsilon}$ so if $|x-y|<\delta$
  • $\left|f(x)-f(y)\right|=\left|x+\frac x{x+1} + y + \frac y{y+1}\right| \le $(triangle inequality) $ |x-y|+\left|\frac x{x+1}-\frac y{y+1}\right| =$ $ |x-y|+\left|\frac {x-y}{(x+1)(y+1)}\right| \le$ $ \delta + \frac \delta{(x+1)(y+1)} \le $ (because $x,y>1$) $ \delta + \frac \delta{4} =\frac 54 \delta = \varepsilon $

I wanted to verify the correctness of this as it took a long time to do it, and it's a first timer for me :-)

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You are correct!
You could also use a result which says that $f$ is uniformly continuous on $F\subseteq \mathbb R$ if $f'(x)$ is bounded on $F$.

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  • $\begingroup$ it's fun to be correct once on a freshmen year :-) $\endgroup$ – jreing Jun 4 '13 at 14:20

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