Proving $\lim_{x\to9}\sqrt{x-5}=2$

Question

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.

Answer 1

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 $\delta < 2\varepsilon$

Answer 2

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} $