Where $f(x) = \frac{x}{x^2+1}$ is continuous On which set is $f(x) = \frac{x}{x^2+1}$ continuous? What method do you use to solve these questions?
 A: Rational functions are continuous everywhere they are defined. In this case, the denominator is never zero, so the function is continuous on the real line.
A: The set of continuous functions are closed under:


*

*addition : $f(x), g(x)$ continuous $\implies f(x) + g(x)$ continuous.

*multiplication: $f(x), g(x)$ continuous $\implies f(x) g(x)$ continuous.

*reciprocal (when it is defined): $f(x)$ continuous $\implies \frac{1}{f(x)}$ continuous whenever $f(x) \ne 0$.


Start with:


*

*any constant (e.g. $x \mapsto 1$ ) is a continuous function.  

*the identity map $x \mapsto x$ is a continuous function.


We have:
$(2) \implies x^2 = x*x$ is continuous.
$(1) \implies x^2 + 1$ is continuous.
$(3) \implies \frac{1}{x^2 + 1}$ is continuous because $x^2 + 1 \ne 0$ for all $x$.
$(2) \implies \frac{x}{x^2 + 1} = x * \frac{1}{x^2+1}$ is continuous.
See the pattern?
A: At first here is a plot of the function

For example we could use the test that $\frac{f}{g}$ with $f$ and $g$ continuous, is continuous when $g\neq 0$.
Beside this we could use the sequence definition of continuousity
$$\lim f(x_n) = \lim_{n \to \infty} \frac{x_n}{x_n^2 +1 } = \frac{\lim x_n}{\lim x_n^2 +1 }= \frac{x}{x^2+1} = f(\lim_{n\to \infty} x_n)$$
If you are interested i can give you more tests
We will formulize the drwaing without removing the pen form the paper definition.
The thing which fits that most is the $\epsilon$-$\delta$ defintion (just ignore the formula I will explain it in words). A function $f$ is continuous in $x_0$ when
$$\forall \varepsilon >0 \, \exists \delta >0\  \text{such that } |x-x_0|< \delta \implies |f(x)-f(x_0)|<\varepsilon$$
you can draw a line when there are no jumps, so that when you go a bit to the right (the left) your function only goes a bit up (or down). This formula says, for every distance of $f(x)$ and $f(x_0)$ you chose, there is always an intervall, around $x_0$ where the values of the function are nearer.
To prove this one here we calculate
$$|f(x)-f(x_0)|=\left| \frac{x}{x^2+1} - \frac{x_0}{x_0^2+1}\right|=\left| \frac{x(x_0^2 +1)-x_0(x^2+1)}{(x_0^2+1) (x^2+1)}\right|$$
Now we say that $x>x_0$, (if not than $-x>-x_0$ and multiplying with $-1$ doesn't matter in the absolute value) so we have
$$\left| \frac{x}{x^2+1} - \frac{x_0}{x_0^2+1}\right|=\left| \frac{x(x_0^2 +1)-x_0(x^2+1)}{(x_0^2+1) (x^2+1)}\right|\leq |x-x_0| \frac{x^2+1}{(x^2+1)(x_0^2+1)}$$
As
$$|x-x_0| \frac{x^2+1}{(x^2+1)(x_0^2+1)}=|x-x_0| \frac{1}{x_0^2+1}\leq |x-x_0|\cdot \frac{1}{2}$$ we see when we move our pencil a bit to the right, we only need to move it $\frac{\text{bit}}{2}$ up (or down).
A: As Jasper Loy said, the "method" for rational functions is to check whether or not the denominator can equal zero. In the case of your example, the denominator cannot be zero, because, for all $x$,  $x^2\ne -1$. In general, if the denominator equals zero, the function is undefined, which means there will be a "hole" in the graph at any $x$ value that makes the denominator zero. (And you will have to lift your pen!)
