Prove that $\frac{1}{\sqrt{1-x^2}}$ is unbounded for $x\in(-1,1)$

Question

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.

Answer 1

$$\frac1{\sqrt{1-x^2}}>M\iff |x|>\sqrt{1-\frac1{M^2}}.$$

This inequation has solutions in $(-1,1)$ for all $M$ ($\ne0$).

Answer 2

Do you really need all that? Let $M>1$. Then from $$ M=\dfrac{1}{\sqrt{1-x^2}} $$ we get $$ x^2=1-\frac{1}{M^2} $$ and this equation has a solution in the interval $(-1,1)$ because $0<1/M^2<1$.

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If you already know that $\lim_{x\to1}f(x)=\infty$ you need nothing more to show that the function is upper unbounded. Indeed, if it were bounded by $K$, then you cannot find (left) neighborhoods of $1$ such that $f(x)>K$.

Answer 3

$\frac{1}{\sqrt{1-x^2}}$ 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.