Proving a Limit of a Square root function using the epsilon-delta definition

Question

Okay, so I've been asked to prove this limit using said epsilon-delta definition:

$ \lim_{x \rightarrow 7} \sqrt{x-3} =2$

I have gotten thus far:

Let epsilon > $0$. Then if delta=_______, then $|x-7|<delta.$

$|f(x)-L|= | \sqrt{x-3} -2|$

$ = |\frac{( \sqrt{x-3}-2)( \sqrt{x-3} +2)}{ \sqrt{x-3} +2} |$

$= |\frac{x-7}{ \sqrt{x-3}+2 } |$

$= \frac{1}{ \sqrt{x-3} +2} |x-7|$

and am now stuck. How do I solve for delta now?

Thanks in advance.

Answer 1

You want to conclude $\frac{|x-7|}{\sqrt{x-3}+2}<\epsilon$ and you can control $|x-7|<\delta$.

We can force $|x-7|\leq 1$ from which follows $x-7\geq -1$ so that $x-3\geq 3$ so that the square root is well defined. Therefore the denominator is bigger than $1$ and so

$$\frac{|x-7|}{\sqrt{x-3}+2}<|x-7|$$

So it is sufficient to pick $\delta\leq \epsilon$. We need to force these two conditions together, so given $\epsilon>0$ we let $\delta=\min(\epsilon, 1)$.