Suppose the function $\sqrt{x}$ is continuous at any $c>0$. Show that this function is differentiable at any $c>0$ and find $f'(c)$.
I know that if a function is continuous any any $c>0$ then for any sequence $x_n \rightarrow c$ we have that $f(x_n) \rightarrow f(c)$
Now I want to show that $\frac{f(x_n)-f(c)}{x_n-c} \rightarrow f'(c)$
So suppose that the function is continuous then let $x_n \rightarrow c$ then we know that $f(x_n) \rightarrow f(c)$ now I can't see how to connect this to what I want to show.
Can I go with trying to evaluate what the limit will be? such as $\lim_{x \rightarrow c} \frac{\sqrt(x_n)-\sqrt(c)}{x_n-c}$ then multiplying by the conjugate $\sqrt(x_n) + \sqrt(c)$ I end up with $\lim_{x \rightarrow c} \frac{1}{\sqrt{x_n}+\sqrt(c)}=\frac{1}{2\sqrt(c)}$
Suppose $f(x)=\sqrt{x}$ is continuous for every $x>0$. Let $c>0$ then $$\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}=\lim_{h\to 0}\frac{\sqrt{c+h}-\sqrt{c}}{h}=\lim_{h\to 0}\frac{c+h-c}{h(\sqrt{c+h}+\sqrt{c})}=\lim_{h\to 0}\frac{1}{\sqrt{c+h}+\sqrt{c}}$$ Since $f$ is continuous then we can pass the limit inside the square root. $$\lim_{h\to 0}\frac{1}{\sqrt{c+h}+\sqrt{c}}=\frac{1}{\lim_{h\to 0}\sqrt{c+h}+\sqrt{c}}=\frac{1}{\sqrt{\lim_{h\to 0}(c+h)}+\sqrt{c}}=\frac{1}{2\sqrt{c}}$$ Exactly similar calculations you can show that $$\lim_{h\to 0}\frac{f(c)-f(c-h)}{h}=\frac{1}{2\sqrt{c}}$$ By definition of the derivative at $c$ we have $$f'(c):=\lim_{h\to 0}\frac{f(c+h)-f(c)}{h}=\lim_{h\to 0}\frac{f(c)-f(c-h)}{h}=\frac{1}{2\sqrt{c}}$$