Prove that the function is bounded: $f(x) = \frac{1}{x^{2}+1}$ 
Prove that the function is bounded: $$f(x) = \frac{1}{x^{2}+1}$$

I'd like to know several (easy) ways of proofing this and I found 2 ways but actually they are so similar, it might just be the same way.. :p


*

*Every convergent function is bounded.

*If function has infimum and supremum then it's bounded. Is it actually bounded if it just has one of both?
1.
$$\lim_{x\rightarrow\infty}\frac{1}{x^{2}+1}= 0$$
$$\lim_{x\rightarrow-\infty}\frac{1}{x^{2}+1}= 0$$
$\Rightarrow 0$ is limit of function so it's convergent and thus bounded.
But what if we had 2 different values, for $\pm\infty$?

2.Calculate supremum:$$\lim_{x\rightarrow0}\frac{1}{x^{2}+1}= \frac{1}{0+1}=1$$
Calculate infimum:
$$\lim_{x\rightarrow\infty}\frac{1}{x^{2}+1}= 0$$ 
Thus function is bounded.

Are both ways correct? If you answer, please answer to all my questions.
 A: "Every convergent function is bounded." - Not quite. You may additionally need that it is continuous on all of $\Bbb R$ (or on its domian and that dmoain  is a closed subset of $\Bbb R$)
"Is it actually bounded if it just has one of both?" -  The function $f(x)=x^2$ has an infimum, but not a supremum. It is bounded from below, but not bounded. Similarly for $f(x)=-x^2$.
"But what if we had 2 different values, for $\pm \infty$" - Doesn't matter. The existence of $\lim_{x\to+\infty}$ shows (together with continuity, see above) that $f$ is bounded on any interval of the form $[a,\infty)$. The existence of $\lim_{x\to-\infty}$ shows (together with continuity, see above) that $f$ is bounded on any interval of the form $(-\infty,a]$. And bounded on both $(-\infty,0]$ and on $[0,\infty)$ implies bounded on $\Bbb R$.
"Calculate Supremum" - You do not make clear that the limits you calculate are in fact the supremum/infimum.
Simpler: $x^2\ge 0$ with eqality iff $x=0$. Hence $x^2+1\ge1>0$ with equality iff $x=0$. Hence $\frac1{x^2+1}\le 1$ with equality iff $x=0$. This shows $\sup f(x)=f(0)=1$ as well as $f(x)>0$, i.e., $\inf f(x)\ge 0$. This already suffices to see that $f$ is bounded, but we can see explicitly that $\inf f(x)=0$ by noting that $\lim_{x\to\infty}f(x)=0$.
A: Setting $x=\tan(a)$ then $f(x)=\frac{1}{1+\tan(a)^2}=\cos(a)^2$ whose range is $[0,1].$
A: If $f(x)\to a\in\mathbb{R^+}$ as $x\to\xi$ then $f$ is bounded in a neighbourhood of $\xi$,, not in general. The fact that $\lim_{x\to\infty}\frac1{x^2+1}=0$ only tells you that the function is asimptotically bounded. You have the same for $1/x$, but it is not bounded on $\mathbb{R^+}$ because it diverges as $x\to0^+$. 
To prove the boundedness of your function, note that since $x^2\ge0$ we have $1+x^2\ge1$ and thus $0<(1+x^2)^{-1}\le1$. Given this inequality, showing that the function is $1$ at $0$ and converges to $0$ as $x\to\pm\infty$ implies that $1$ is the maximum (so the supremum) and $0$ the infimum.
By the way, $\inf x^{-2}=0$ but $\sup x^{-2}=+\infty.$
A: HINT.-Just to give another way.
Never $f(x)\lt 0$. If for $N$ large we have $f(x)\gt N$ then $$\frac{1}{x^2 + 1}\gt N\iff\frac 1N\gt x^2+1$$ It follows that LHS tends to $0$ and RHS tends to $\infty$, absurde.
A: The function $f(x) := x^2 + 1$ is differentiable (since it is a polynomial) $\Longrightarrow f'(x) = 2x$. This derivative will vanish if and only if $x = 0$, hence $f$ takes its minimum (it can only be a minimum by graphical considerations!) at $x = 0$ with $f(0) = 1$. Therefore, you have $\max \limits_{x \in \mathbb{R}} \frac{1}{x^2 + 1} = \frac{1}{\min \limits_{x \in \mathbb{R}} f(x)} = \frac{1}{1} = 1 \Longrightarrow f$ is bounded from above. To see that $f$ is bounded from below, it suffices to check that both, numerator and denominator, are strictly positive functions, hence the function $\frac{1}{x^2 + 1}$ is bounded from below by zero.
