I have to prove that $$\lim_{x \to a} \frac{1}{\sqrt{x}}=\frac{1}{\sqrt{a}}$$ where x>0 and a>0.
So given $\epsilon>0$, I let $\delta<a/2$. Then:
$$|x-a|<\delta \\ -\delta<x-a<\delta \\ a-\delta<x<a+\delta \\ a/2<x<3a/2$$
So now
$$|1/(\sqrt{x})-1/(\sqrt{a})| =|(\sqrt{a}-\sqrt{x})/(\sqrt{xa})|\\ =|(\sqrt{a}-\sqrt{x})(\sqrt{a}+\sqrt{x})/(\sqrt{xc}(\sqrt{a}+\sqrt{x}))| \\ =|(x-a)/(\sqrt{xa}(\sqrt{a}+\sqrt{x}))| \\ <|(x-a)/(\sqrt{xa}(\sqrt{a})| \\ <\delta/(\sqrt{a/2*a}\sqrt{a})\\ <2\delta/(a^{3/2})<\epsilon$$
so given any $\epsilon>0$, let $\delta<min(a/2, \epsilon*a^{3/2}/2)$
so now if $\delta<\epsilon*a^{3/2}/2<a/2$
$|1/(\sqrt{x})-1/(\sqrt{a})|<2\delta/(a^{3/2})<2\epsilon*a^{3/2}/(2a^{3/2})=\epsilon$,
and now if $\delta<a/2<\epsilon*a^{3/2}/2$, then
$|1/(\sqrt{x})-1/(\sqrt{a})|<2\delta/(a^{3/2})<2(a/2)/a^{3/2}<2\epsilon*a^{3/2}/(2a^{3/2})=\epsilon$
Does this proof work? Thanks
It does work!
However, you should learn to type by Tex, for example $a/2$, should be written that ${{\frac{a}{2}}}$ by using "\frac", * is written $\times$ by using "\times" .