A multiple choice question on a function Let $f(x) = x^2+ {1\over x^2}$ for $x \in (0,\infty)$, then which of the following is/are correct:
(a) $f$ is continuous on $(0,\infty)$.
(b) $f$ is uniformly continuous on $(0,\infty)$.
(c) $f$ attains its supremum on $(0,\infty)$.
(d) $f$ attains its infimum on $(0,\infty)$.
Would like to share my thoughts on the same!
I know that if $f$ comes out to be uniformly continuous then it will be continuous as well, so working with option (b) first will consequently save my time.
To proceed I was in search of $a_n$ and $b_n$ such that $|a_n - b_n| \to 0$ but $|f(a_n) - f(b_n)| \not\to 0$. By choosing $a_n = {1\over n}$ and $b_n = {2\over n}$  I concluded that $|f({1\over n}) - f({2\over n})| = {3n^2\over 4}-{3\over n^2} \not\to 0$. So option b false.
Now separately I'll have to check for option (a),(c) and (d) as well. Graphically my function looks like:
that partially helped me to answer the question! Is there any theoretical way to get the conclusion?
Thanks in advance :).
 A: Note that the function is symmetric along y-axis (as $x^2$ is the only term used). The function cannot have a supremum if, for any number k, the inequality:
\begin{align}x^2+\frac{1}{x^2}\ge k\end{align}
Which can be stated as:\begin{align}x^4-kx^2+1\ge 0\end{align}
Has a solution. To see this, firstly, equality holds when $x^4-kx^2=-1$, and solutions are $-\sqrt{\frac{k-\sqrt{k^2-4}}{2}},\sqrt{\frac{k-\sqrt{k^2-4}}{2}},-\sqrt{\frac{k+\sqrt{k^2-4}}{2}},\sqrt{\frac{k+\sqrt{k^2-4}}{2}}$
The solutions exist only for $k\ge 2$, But if for any $k\ge 2$, the inequality holds, it automatically does for $k<2$.
The derivative of the function $x^4-kx^2+1$ is
\begin{align}4x^3-2kx=2x^2+2x(x^2-k)=2x^2-\frac{2}{x}\end{align}
As $x\ge 1$ in the fourth solution (the largest one), it follows that the derivative is also $\ge 0$.
Thus, the inequality has a solution.
The infimum exists as the derivative is zero at 1 and -1, and the function does not have a supremum.
To analyse uniform continuity, take $(x_1,x_2)$ in the interval $(0,\infty)$ so that $x_2-x_1=\delta$, then:
\begin{align}\biggr|x_1^2+\frac{1}{x_1^2}-x_2^2-\frac{1}{x_2^2}\biggr|=|x_1^2-x_2^2|\biggr|1-\frac{1}{x_1^2x_2^2}\biggr|= \delta|x_1+x_2|\biggr|1-\frac{1}{x_1^2x_2^2}\biggr|<\epsilon\end{align}
For $x_1,x_2\rightarrow \infty$, the coefficient of $\delta$ increases. For an arbitrary constant value of $\delta$
\begin{align}\delta|x_1+x_2|\biggr|1-\frac{1}{x_1^2x_2^2}\biggr|>\epsilon\end{align}
For any large enough $x_1,x_2$.
So the function is not uniformly continuous. Continuity is established by dividing both the sides by the coefficient of $\delta$
