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.
$$\lim_{x \to a} \frac{f(x)-f(a)}{\sqrt{x}-\sqrt{a}}$$
$$=\lim_{x \to a} \frac{f(x)-f(a)}{\sqrt{x}-\sqrt{a}}\cdot \frac{\sqrt{x}+\sqrt{a}}{\sqrt{x}+\sqrt{a}}$$
$$=\lim_{x \to a} \frac{(f(x)-f(a))(\sqrt{x}+\sqrt{a})}{x-a}$$
$$=\lim_{x \to a} \left(\frac{f(x)-f(a)}{x-a}\cdot (\sqrt{x}+\sqrt{a})\right)$$
$$=\lim_{x \to a} \frac{f(x)-f(a)}{x-a}\cdot \lim_{x \to a}(\sqrt{x}+\sqrt{a})$$
$$=f'(a)\cdot \lim_{x \to a}(\sqrt{x}+\sqrt{a})$$
$$=1\cdot \lim_{x \to a}(\sqrt{x}+\sqrt{a})$$
$$=\lim_{x \to a}(\sqrt{x}+\sqrt{a})$$
$$=\sqrt{a}+\sqrt{a}$$
$$=2\sqrt{a}$$

You made a mistake. (It's not a big one but it impacts your answer).
On Line $3$, you should have had $\sqrt{x}+\sqrt{a}$ in numerator.
Now, redo the steps (easy), and you'll end up with $2\sqrt{a}$. 
