# Question about derivative of $\sqrt{x}$ at $x=9$.

I am asked the following: Suppose you know that the derivative of $$\sqrt{x}$$ is $$\dfrac{1}{2\sqrt{x}}$$ for every $$x >0$$. Then

$$\lim_{x \to 9} \dfrac{\sqrt{x}-3}{x-9} = \dfrac{1}{a}$$

where $$a=6$$.

In this problem, are we just supposed to plug-in $$9$$ into

$$\dfrac{1}{2\sqrt{x}}$$ by acknowledging that the definition of derivative of $$\sqrt{x}$$ at $$9$$ is $$\lim_{x \to 9} \dfrac{\sqrt{x}-3}{x-9}$$ and so is equal to $$\dfrac{1}{2\sqrt{x}}$$ at $$x = 9$$?

• If you were told to use the derivative fact, then use it. Otherwise, calculate the limit using algebra and properties of limits. Commented Sep 5 at 3:39
• It depends on the question context. Sometime you are asked to use this to find by definition the derivative of $\sqrt{x}$ at $x=9$. However, in other categories especially complex variable stuff, it is useful by identifying the limit is the derivative of some known functions. Commented Sep 5 at 4:20

In this problem, are we just supposed to plug-in $$9$$ into $$\dfrac{1}{2\sqrt{x}}$$ by acknowledging that the definition of derivative of $$\sqrt{x}$$ at $$9$$ is $$\lim_{x \to 9} \dfrac{\sqrt{x}-3}{x-9}$$ and so is equal to $$\dfrac{1}{2\sqrt{x}}$$ at $$x = 9$$?
Yes, exactly. More broadly, $$f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}$$ Granted, this is an "easy way" to figure out the limit, if you already know the derivative.
Mathematically, rigorously, it's a little less kosher since this limit is (usually) needed to figure out said derivative to begin with, so usually you need to appeal to methods that don't use derivatives. For instance, $$\frac{\sqrt x - 3}{x- 9} = \frac{\sqrt x - 3}{x- 9} \cdot \frac{\sqrt x + 3}{\sqrt x + 3} = \frac{x-9}{(x-9)(\sqrt x + 3)} = \frac{1}{\sqrt x + 3}$$ and then the limit is obvious.