Proving $\lim_{x\to9}\sqrt{x-5}=2$ Im stuck on proving $\displaystyle \lim_{x\to9}\sqrt{x-5}=2.$
I start with let $e>0$ be given we need to find a $d>0$ such that whenever
$0<|x-9|<d$ we have $|f(x)-2|<e$
My method is to start with $|f(x)-2|<e$ isolate $x$ and then i think i end up with an unsymetrical interval about $9$ so i choose $d>0$ to be the distance to the shortest endpoint. Once i get my delta i attempt to show the implication but can't. 
i expanded that inequality out to get
$9-4e+e^2 < x < 9 + 4e + e^2$
and from here i took $d$ to be $-e^2 + 4e$ and i can't show the impllication. Could someone tell me if i have done something wrong?
i also think i assumed $e<2$ during that process, is it ok to do this as becuase we can just use the delta given by say $e=1$ whenever we choose $e>2$?
Thanks for your help.
 A: Let $\varepsilon > 0$,
$|\sqrt{x-5} - 2| = |(\sqrt{x-5} - 2) \frac{\sqrt{x-5}+2}{\sqrt{x-5}+2}| = |\frac{(\sqrt{x-5})^{2}-4}{\sqrt{x-5}+2}| = \frac{|x-5-4|}{|\sqrt{x-5}+2|} = \frac{|x-9|}{\sqrt{x-5}+2} \leq \frac{|x-9|}{2} < \frac{\delta}{2} < \varepsilon $
I use that $\sqrt{x-5} > 0 \implies$ $\sqrt{x-5} + 2 > 2 \implies$ $\frac{1}{\sqrt{x-5} + 2} < \frac{1}{2}$
And I take $\delta < 2\varepsilon$
A: An important thing
to remember for limits
is that you do not need
the best $\delta$ -
any $\delta$ will do.
You want to show that
$\lim_{x \to 9}\sqrt{x-5}=2
$.
Another of my preferences
is to always let variables
approach zero.
In this case,
let $x = y+9$.
$\lim_{x \to 9}\sqrt{x-5}
$
becomes
$\lim_{y \to 0}\sqrt{y+9-5}
=\lim_{y \to 0}\sqrt{y+4}
$.
A simple way
to handle this is
$\begin{array}\\
\sqrt{y+4}-2
&=(\sqrt{y+4}-2)\dfrac{\sqrt{y+4}+2}{\sqrt{y+4}+2}\\
&=\dfrac{(y+4)-4}{\sqrt{y+4}+2}\\
&=\dfrac{y}{\sqrt{y+4}+2}\\
&\text{so}\\
|\sqrt{y+4}-2|
&=\dfrac{|y|}{\sqrt{y+4}+2}\\
&<\dfrac{|y|}{2}
\qquad\text{if } y > -4\\
\end{array}
$
