Continuity of function

[man111]

Discuss the continuity of [tex]f:\mathbb{R^{+}}\rightarrow \mathbb{R}[/tex] defined as [tex]f(x) = x,[/tex] When [tex]x[/tex] is Irrational

and [tex]\displaystyle f(x) = \left(\frac{1+p^2}{1+q^2}\right)^{\frac{1}{2}},[/tex] When [tex]x[/tex] is rational number of the form [tex]\displaystyle \frac{p}{q}[/tex]

[nevrodermit]

Let [tex](x_n)[/tex] is a sequence of rational that tends to the irrational number [tex]x[/tex]. Let the function is continuous at [tex]x[/tex]. Then [tex]f(\lim(x_n))=f(x)=x[/tex]. Let [tex]x_n=\frac{p_n}{q_n}[/tex], then [tex]\lim_{n\to \infty}{\sqrt{\frac{1+p_n^2}{1+q_n^2}}}=x[/tex].
So [tex]p_n \to xq_n[/tex] and [tex]1+p_n^2\to x^2(1+q_n^2)[/tex] and thus [tex]1+x^2q_n^2 \to x^2+x^2q_n^2[/tex] and this [tex]x=1[/tex]. So no continuity at irrationals.

Let [tex](x_n)[/tex] is a sequence of irrationals that tends to the rational number [tex]x=\frac{p}{q}[/tex]. Let the function is continuous at [tex]x[/tex]. Then [tex]\frac{p}{q}=\lim{(x_n)}=\lim{f(x_n)}=f(\lim{(x_n)})=f(x)=\sqrt{\frac{1+p^2}{1+q^2}}[/tex]. So [tex]\frac{p^2}{q^2}=\frac{1+p^2}{1+q^2} \Leftrightarrow p=\pm q[/tex]. So it is continuous at [tex]x=1[/tex]

[drago]

So [tex]p_n \to xq_n[/tex] and [tex]1+p_n^2\to x^2(1+q_n^2)[/tex]

No, you cannot write things like that. The both sides tend to $\infty$

So, let $x$ is irrational and $p_n/q_n\to x$. Note that $p_n\to \infty,q_n\to \infty$. Then $\sqrt{\frac{1+p_n^2}{1+q_n^2}}=\frac{p_n}{q_n}\sqrt{\frac{1+1/p_n^2}{1+1/q_n^2}}\to x$. It means $f$ is continuous at irrationals.

[nevrodermit]

Yes, you are correct.