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 $x = 9$?
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.
