If $f:[0,2]\to\Bbb R$ be defined by $f(x)=\frac {1}{x^2 -2}$. Is f uniformly continuous.
I know continuous function on a closed and bounded interval is uniformly continuous. But here it is not so. Because as $n \to \sqrt2 , the function f(x)$ goes to $\infty$.
where does my concept goes wrong ?
Your function is defined at $$D=[0,\sqrt {2}[\cup ]\sqrt {2},2] $$
take $$x_n=\sqrt {2}+\frac {1}{n}$$ $x_n\in D $ and $(x_n) $ is Cauchy.
but
$f (x_n)=\frac {n^2 }{ 1+2n\sqrt {2}}$
is not Cauchy cause it diverges. So $f $ is not uniformly continuous at $D $.