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My goal is to prove $f(x)=\frac{1}{\sqrt{1+x^2}}$ is continuous. I think it will be easier to approach this using the def of sequential continuity?

Proof: Prove $\left\{a_{n}\right\}$ converges to a then $\left\{\sqrt{a_{n}}\right\}$ converges to $\sqrt{a}$ : suppose $\left\{a_{n}\right\}$ converges to $a, \lim \left(a_{n}-a\right)=$ $\lim \left(a_{n}^{\frac{1}{2}}-a^{\frac{1}{2}}\right) \lim \left(a_{n}^{\frac{1}{2}}+a^{\frac{1}{2}}\right)=0$ Thus when $a \neq 0,\left\{\sqrt{a_{n}}\right\} \rightarrow a^{\frac{1}{2}}$ Define $g(x)=\frac{1}{1+x^{2}}$. since $g(x)$ is continuous according to Theoren $3.4,$ Thus $\forall x_{0} \in R,$ whenever $\left\{x_{n}\right\} \rightarrow x_{0},\left\{g\left(x_{n}\right)\right\} \rightarrow g\left(x_{0}\right) .$ Thus $,\left\{\sqrt{g\left(x_{n}\right)}\right\} \rightarrow \sqrt{g\left(x_{0}\right)} .$ Thus $\forall x_{0} \in R,$ whencrer $\left\{x_{n}\right\} \rightarrow$ $x_{0},\left\{f\left(x_{n}\right)\right\} \rightarrow f\left(x_{0}\right) .$ Thus $f$ is continuous.

Is this correct?

edit:$x\in(0,\infty)$

catitpillar
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    What's "theorem 3.4"? – 5xum Oct 26 '20 at 07:47
  • @5xum that's on Fitzpatrick, basically, polynomial function is continuous and if f,g are continuous, f/g is continuous – catitpillar Oct 26 '20 at 07:49
  • If you already know that polynomials, $\frac{1}{x}$ and $\sqrt{x}$ are continuous wherever they are defined, then your function is a composition of those functions. Thus continuous. – freakish Oct 26 '20 at 07:51
  • @freakish but I don't know $\sqrt{x}$ is continuous right? polynomials need to have an integer degree? – catitpillar Oct 26 '20 at 07:53
  • @catitpillar if you don't know that $\sqrt{x}$ is continuous then prove it first: https://math.stackexchange.com/questions/560307/prove-that-sqrtx-is-continuous-on-its-domain-0-infty And no: typically the standard proof about polynomials being continuous is not applicable to non-natural powers. – freakish Oct 26 '20 at 07:55
  • This function is continuous in the whole real line – Shreya Chauhan Oct 26 '20 at 07:56

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