Proof that $\;\dfrac{x}{1+x^2}\;$ is a bounded function. I don't quite understand how $\dfrac1 x$ is unbounded while $\dfrac{x}{1+x^2}$ is not.
 A: Look at where $\dfrac 1 x$ goes to infinity,  what causes that unbounded behavior?  Notice that that is avoided in the second function.
A: Here it is an alternative approach for the sake of curiosity.
\begin{align*}
|x^{2} + 1| = x^{2} + 1 \geq 2\sqrt{x^{2}} = 2|x| \Longleftrightarrow \left|\frac{x}{x^{2}+1}\right| \leq \frac{1}{2} \Longleftrightarrow -\frac{1}{2} \leq \frac{x}{x^{2}+1}\leq\frac{1}{2}
\end{align*}
and we are done.
Hopefully this helps!
A: Regarding $x/(1+x^2)$, it may help to analyze the cases $|x| \leq 1$ and $|x| > 1$ separately.
In the former case you have a continuous function everywhere defined on a closed interval (here it's important that there's squaring in the denominator; nothing is getting close to zero down there), so it's going to be bounded. You can also do algebra with the hypothesis $|x| \leq 1$ and derive an explicit bound, or be more sophisticated and verify that a particular number is the least upper bound for the function on that interval.
For $|x| > 1$ one can leverage the fact that in this instance $|x| < |x|^2$ and hence $|x/(1 + x^2)| = |x|/(1 + |x|^2) \leq |x|^2/(1 + |x|^2)$ and this last fraction is less than $1$ because it has the form $a/b$ with $a$ and $b$ positive numbers and $a < b$.
A: We have
$$\quad\lim\limits_{x \to 0}\space (x)=0
\qquad \text{while}\qquad 
\lim\limits_{x \to 0}\space (x^n+1)=1\quad$$
so the limit is as follows
$$
\lim\limits_{x \to 0}\frac{1}{x}=\infty\qquad
\text{while}\qquad 
\lim\limits_{x \to 0}\bigg(\frac{x}{x^2+1}\bigg)
=\left(\frac{\lim\limits_{x \to 0}\space (x)}
       {\lim\limits_{x \to 0}\space (x^n+1)}\right)
=\frac{0}{1}=0$$
