I'm trying to use the $\epsilon-\delta$ definition of limits to show
$$\lim_{x\to p} \sqrt{x}=\sqrt{p}$$ assuming $p>0$.
I know that $|\sqrt{x}-\sqrt{p}|<\epsilon$ whenever $0<|x-p|<\delta$
Then after some algebra I get $\frac{|x-p|}{|\sqrt{x}+\sqrt{p}|}<\epsilon$
But I am lost as what to do next. I know I need to end up having something which looks like $|x-p| < \epsilon$.
Any help in the right direction would be great! Thanks.
You are almost there.
Well first of all we do not know that $|\sqrt{x}-\sqrt{p}|<\epsilon$ whenever $0<|x-p|<\delta$. This is what we are required to prove. As long as we assume that the function $f$ is defined only for $x \ge 0$. Then since $|\sqrt{x}+\sqrt{p}| \ge \sqrt{p} \ge 0:$
$$|\sqrt{x}-\sqrt{p}| \le \frac{|x-p|}{|\sqrt{x}+\sqrt{p}|} \le \frac{|x-p|}{\sqrt{p}}$$
Now to the logical side of things.
What you are trying to prove is that $\forall \epsilon \gt 0,\;\;|\sqrt{x}-\sqrt{p}|<\epsilon \implies |x-p| \lt \delta$ for some $\delta \gt 0$
What we need to prove instead is "Given any $\epsilon \gt 0$ however small, $\exists \delta \gt 0$ such that $|x-p| \lt \delta \implies |\sqrt{x}-\sqrt{p}|<\epsilon $";
Now let $\epsilon \gt 0$ be an arbitrary positive quantity thrown at us. We know that $|\sqrt{x}-\sqrt{p}| \le \frac{|x-p|}{\sqrt{p}}$ so we can say with confidence that if $ |x-p| \lt \sqrt p \cdot \epsilon $ then surely $|\sqrt{x}-\sqrt{p}| \lt \epsilon$. Here the $\delta$ supplied by us is $\sqrt p \cdot \epsilon $ and we shown that it works for any $\epsilon \gt 0$. And you're done..
Alternatively you can also proof the assertion by sequences.
(1) Suppose that $p=1$, then we'd like to show that $\lim_{x\rightarrow 1; x\in (0,\infty)}x^{1/2}=1$. Let $(a_n)_{n=0}^\infty$ be a sequence of elements in $(0,\infty)$, such that $(a_n)\rightarrow 1$. Then we will show that $(a_n^{1/2})\rightarrow 1$.
Given $\varepsilon>0$ it will suffice if we can show that $(a_n^{1/2})$ is $\varepsilon$-close to $1$. We already know that $(1+1/n)\rightarrow 1$ and $(1-1/n)\rightarrow 1$. Let $K$ be such that both are $\varepsilon$-close to $1$. Now since $(a_n)\rightarrow 1$, then there is some $N$ such that $|a_n-1|\le 1/K$ for all $n\ge N$. So
$$1-1/K\le a_n\le1+1/K$$ $$(1-1/K)^{1/2}\le a_n^{1/2}\le(1+1/K)^{1/2}$$
Since $1-1/K<1$, then $(1-1/K)^{1/2}>(1-1/K)$ and similarly since $1+1/K>1$ so $(1+1/K)^{1/2}<(1+1/K)$. Thus
$$(1-1/K)\le a_n^{1/2}\le(1+1/K)$$
and hence $a_n^{1/2}$ is $\varepsilon$- close to $1$ as desired.
(2) Let $p \in (0 \infty)\backslash\{1\}$ and let $(p_n)_{n=0}^\infty$ be a sequence of positive real numbers which converges to $p$. Using the limit laws we know that $p_n/p$ converges to $1$ and by part $1)$, we have $(p_n/p)^{1/2} \rightarrow 1$. Thus
\begin{align}\lim_{n\rightarrow \infty}p_n^{1/2}=\lim_{n\rightarrow \infty} p^{1/2} (p_n/p)^{1/2}\\ = p^{1/2} \lim_{n\rightarrow \infty} (p_n/p)^{1/2} \end{align}
and since $\lim_{n\rightarrow \infty} (p_n/p)^{1/2}=1 $, hence $\lim_{n\rightarrow \infty}p_n^{1/2}=p^{1/2}$. Since $(p_n)$ was an arbitrary sequence of positive real numbers converging to $p$, this would imply $\lim_{x\rightarrow p: x\in (0,\infty)} x^{1/2}=p^{1/2}$.
