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Here is my attempt.

So starting with $f(x_0)=f(1)=1$ and $\epsilon >0$, I want to find $\delta >0$ such that if $|x-1|< \delta$ I get $|f(x)-f(x_0)| \leq \epsilon$.

I started studying $|f(x)-f(x_0)|$ as follows

$|f(x)-f(x_0)|= |\frac{x^2+1}{x+1}-1|= |\frac{x^2-x}{x+1}|= |x \frac{x-1}{x+1}| \leq |\frac{x}{x+1}| \delta$

But now I cannot find a way to conclude. I want to find an upper bound for $|\frac{x}{x+1}|$ and then find an expression for $\delta$ in terms of $\epsilon$

$|f(x)-f(x_0)|= |\frac{x}{x+1}| \delta \leq \frac{(\delta +1)\delta}{2- \delta}$

From now on I tried to set $ \frac{(\delta +1)\delta}{2- \delta} = \epsilon$ and solve for $\delta$ but this didn't work.

I then noticed that $\frac{(\delta +1)\delta}{2- \delta} \leq \frac{(\delta +1)^2}{2- \delta}$ from where I tried solving $\frac{(\delta +1)^2}{2- \delta}=\epsilon$ but again, I failed to find an applicable solution.

I have the feeling that I am near but somehow I cannot see how to overcome the difficulty.

Thank you for your help.

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  • $\begingroup$ try writing x as 1+dx. Now in your expression for |f(x) - f(x0)| all you have to do is show that if dx approaches 0 the expression does aswell. $\endgroup$ – Jagol95 Dec 16 '18 at 20:32
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Unfortunately, $$\frac{x}{x+1}=1-\frac1{x+1}$$ is an unbounded function, so that $\left\lvert\frac{x}{x+1}\right\rvert$ has no upper bound. However, we can bound it by making sure that $x$ is bounded away from $-1,$ at which the function has its vertical asymptote. Fortunately, this isn't an issue, since we want to keep $x$ near $1,$ anyway.

So long as $x$ is positive, $\frac{x}{x+1}$ will be positive, as well, and necessarily less than $1.$ Can you see why?

Consequently, picking some arbitrary $\alpha\in(0,1),$ we need only make sure that $\delta=\min\{\alpha,\epsilon\},$ at which point we'll have $$\left\lvert\frac{x}{x+1}\right\rvert\delta<\delta\le\epsilon$$ whenever $|x-1|<\delta,$ which completes the proof.

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  • $\begingroup$ thank you for your comment. It was really helpful. As you may have seen, I tried to solve my problem by setting $\delta = min${$\epsilon;\frac{1}{2}$}. Do you think this will also do the job ? I choose suche a $\delta$ because $|\frac{x}{x+1}| \leq \frac{\delta +1}{2- \delta}$ and so I tried to solve $\frac{\delta +1}{2- \delta} \leq 1$ and found that $\delta$ should be in $(- \infty ; \frac{1}{2}] \bigcup (2; \infty)$ $\endgroup$ – Alain Dec 16 '18 at 21:51
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    $\begingroup$ That absolutely works. Any $\alpha\in(0,1)$ will do the job, including $\alpha=\frac12.$ $\endgroup$ – Cameron Buie Dec 16 '18 at 21:52
  • $\begingroup$ Many thanks for your kind help $\endgroup$ – Alain Dec 16 '18 at 21:53
  • $\begingroup$ +1: Well and helpfully explained. :-) $\endgroup$ – user445909 Dec 16 '18 at 23:53
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You could choose a $\delta$ which ensures both $|\frac{x}{x+1}| < 1$ and $\delta < \epsilon$ so you can easily extend the steps in your first line of workings.

It can be shown that $|\frac{x}{x+1}| < 1 \ \text{if and only if} \ x > -\frac 12$, and we know $ 0 < |x-1| < \delta \iff 1 - \delta < x < 1 + \delta$ and $x \neq 1$. Therefore, we want a $\delta$ such that $1 - \delta \geq -\frac 12 \iff \delta \leq \frac 32$.

We have narrowed down our requirements for $\delta$ to

  • $\delta \leq \frac 32$
  • $\delta < \epsilon$
  • $\delta > 0$

I would then suggest finding a function of a real positive variable which satisfies these conditions.

An example of such $\delta$ is $\delta = \frac{\epsilon}{\epsilon+1}$.

$\textbf{Edit}$: It is fine to relax the condition $\delta < \epsilon$ to $\delta \leq \epsilon$ as shown in Cameron's answer.

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  • $\begingroup$ +1: Very elegant! I like it. $\endgroup$ – Cameron Buie Dec 16 '18 at 21:49
  • $\begingroup$ Thank you ! That's a very clever way of doing it. $\endgroup$ – Alain Dec 16 '18 at 21:53

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