# Proof using derivative information to find limit

This is the last exercise of a quite challenging exercises paper a friend who is taking calculus has which I'm trying to help. I already helped her doing the other bunch. But this got me. I will appreciate anyone help to see my work and to tell me if is right or If I need to correct something.

The exercise is:

If $$f'(a)=1$$ for $$a>0$$, find $$\lim_{x \to a} \frac{f(x)-f(a)}{\sqrt{x}-\sqrt{a}}$$.

What came to my mind was to rationalize the denominator.

$$\lim_{x \to a} \frac{f(x)-f(a)}{\sqrt{x}-\sqrt{a}}$$

$$=\lim_{x \to a} \frac{f(x)-f(a)}{\sqrt{x}-\sqrt{a}}\cdot \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}$$

$$=\lim_{x \to a} \frac{(f(x)-f(a))(\sqrt{x}+\sqrt{a})}{x-a}$$

$$=\lim_{x \to a} \left(\frac{f(x)-f(a)}{x-a}\cdot (\sqrt{x}+\sqrt{a})\right)$$

$$=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}\cdot \lim_{x \to a}(\sqrt{x}+\sqrt{a})$$

$$=f'(a)\cdot \lim_{x \to a}(\sqrt{x}+\sqrt{a})$$

$$=1\cdot \lim_{x \to a}(\sqrt{x}+\sqrt{a})$$

$$=\lim_{x \to a}(\sqrt{x}+\sqrt{a})$$

$$=\sqrt{a}+\sqrt{a}$$

$$=2\sqrt{a}$$

On Line $$3$$, you should have had $$\sqrt{x}+\sqrt{a}$$ in numerator.
Now, redo the steps (easy), and you'll end up with $$2\sqrt{a}$$.