# show that the function $f(x)= \frac{x^2+1}{x+1}$ is continuous at the point $x_0=1$

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.

• 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. – Jagol95 Dec 16 '18 at 20:32

## 2 Answers

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.

• 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 $(- \infty ; \frac{1}{2}] \bigcup (2; \infty)$ – Alain Dec 16 '18 at 21:51
• That absolutely works. Any $\alpha\in(0,1)$ will do the job, including $\alpha=\frac12.$ – Cameron Buie Dec 16 '18 at 21:52
• Many thanks for your kind help – Alain Dec 16 '18 at 21:53
• +1: Well and helpfully explained. :-) – user445909 Dec 16 '18 at 23:53

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.

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