# Proof that in $\mathbb{R}^+$ the function $\frac{1}{1+x}$ is continuous

I want to proof that the function $$\frac{1}{1+x}$$ is continuous in the positive real numbers. Therefore I use the epsilon delta definition of continuity. But when I make my calculations (to find epsilon in function of delta) I get stuck.

Can anybody help me to go further on $$\left| \frac{x-a}{(1-x)(1-a)} \right| < \epsilon$$.

• Why do you have $(1-x)(1-a)$ in the denominator? Feb 25, 2021 at 19:12
• Because according to the defintion |f(x)-f(a)| u have $| \frac{1}{1-x} - \frac{1}{1-a}|$ and then make them have the same denominator. Feb 25, 2021 at 19:24
• except the function is $\frac1{1+x},$ not $\frac1{1-x}.$ Feb 25, 2021 at 19:26
• Ow sorry quick mistake! Feb 25, 2021 at 19:30

It should read: $$\left|\dfrac{x-a}{(1+x)(1+a)}\right| < |x-a| < \epsilon$$ if you let $$\delta = \epsilon > 0$$.