Prove that $\frac{1}{\sqrt{1-x^2}}$ is unbounded for $x\in(-1,1)$ I'm trying to prove that $f(x):=\frac{1}{\sqrt{1-x^2}}$ is unbounded for $x\in(-1,1)$; so I must prove that for all $M\in\mathbb{R}$ is $f(x) > M$ for $x\in(-1,1)$.
With the definition of limit, since
$$\lim_{x \to 1^-} f(x)=\lim_{x \to -1^+} f(x)=\infty$$
For all $M>0$ exists $\delta_M>0$ such that if (for instance) $1-\delta_M<x<1$ then $f(x)>M$, so since $M>0$ is arbitrary the definition of unbounded function is satisfied; the doubt is that I don't know how to work with the $x$ interval for which that is true, I mean that I know from the definition of limit that when $1-\delta_M < x <1$ it is $f(x)>M$ but how do I make this rigorous?
I was thinking something like this: since $f$ is continuous in $(-1,1)$, for the Weierstrass theorem $f$ has maximum and minimum (so is bounded) in every interval of the kind $[-1+t,1-t]$ for $t>0$; so if I choose $\delta_M<t$ I can conclude that $f$ is unbounded because it is greater than $M$ for all $M\in\mathbb{R}$ in the interval $[1-\delta_M,1) \subset [1-t,1)$. Is this correct?
However if this is correct I have another doubt: how do I prove that $[-1+t,1-t]$ is limited and closed? I know it is trivial, but I've never done this before.
Another attempt is by contradiction: suppose that $f$ is bounded, then exists $M\in\mathbb{R}$ such that $|f(x)| \leq M$ for all $x\in(-1,1)$; so we have
$$0<\frac{1}{\sqrt{1-x^2}} \leq C \Leftrightarrow0<1\leq C\sqrt{1-x^2}$$
Taking the limit for $x \to1^-$ both sides we have that
$$0\leq \lim_{x \to 1^-} 1 \leq \lim_{x \to 1^-} C\sqrt{1-x^2}=0$$
So by comparison we have
$$\lim_{x \to 1^-} 1=0$$
Which is a contradiction. Is this correct?
Thanks.

$\frac{1}{\sqrt{1-x^2}}$ is an even function, it suffices to consider $x \in [0,1).$
Assume $(1-x^2)^{1/2} <M (>0)$, real, for all $x \in [0,1)$.
Then $1/M^2 < 1-x^2$;
Set $y_n:=x_n^2=1-1/n$;
$1/M^2 < 1/n$, for large enough $n$ a contradiction.
