Prove $f:D \to \mathbb{R}, x \to x+\frac{1}{x}$ is continuous on $D:= [2,13]$ Let $D:= [2,13]$ and $f:D \to \mathbb{R}, x \to x+\frac{1}{x}$
How can one prove that $f$ is continuous on $D$ ?
I know that the epsilon-delta-criterion states that $f$ is continuous in $x_0$ if for every $\epsilon > 0$ there exists a $\delta > 0$ such that for all $x \in D_f$ with $|x-x_0| < \delta$ it holds that $|f(x)-f(x_0)| < \epsilon$.
The problem I have is that $x_0$ in this case is the interval $[2,13]$.
I think that we need an epsilon neighborhood $(x_0 - \epsilon, x_0 + \epsilon)$ and then $x_0 + \epsilon \leq 2$ and $x_0 - \epsilon \geq 13$.
I would start with
$$\big | x+\frac{1}{x} - x_0 - \frac{1}{x_0} \big | < \epsilon \iff ?$$
but then ?
 A: Hint:
For $x,x_0$ in $[2,13]$,  $$\left|x+\frac{1}{x} - x_0 - \frac{1}{x_0}\right|\le\left|x-x_0\right|\left|1+\frac1{x\,x_0}\right|\le\left|x-x_0\right|\frac54.$$
A: So let's choose an $\epsilon > 0$. The goal is to find a positive $\delta_\epsilon$ such that $|f(x)-f(x_0)| < \epsilon$.
Using the triangular inequality one have, $$\big | x+\frac{1}{x} - x_0 - \frac{1}{x_0} \big | \leq \big | x - x_0 \big | + \big |  \frac{1}{x}- \frac{1}{x_0} \big |$$
Now try to find an expression for a $\delta_1$ (that depends on epsilon) that satisfies $| x - x_0 \big | < \frac{\epsilon}{2}$ and another for $\delta_2$ (that depends on epsilon) that satisfies $  \big |  \frac{1}{x}- \frac{1}{x_0} \big |  < \frac{\epsilon}{2}$ then take $\delta_\epsilon = min(\delta_1, \delta_2)$. Hence for any $\epsilon > 0$ you have  a $\delta_\epsilon$ satisfying the inequality.
Basically we are proving a special case of the property that the sum of two continuous functions is continuous.
Tell me if that is not clear enough.
