How do you evaluate the limit of
$$\lim_{x \to 9} \frac {x-9} {\sqrt{x} – 3}$$
by first principle? I was thinking of this way to evaluate but it seems to evaluate to $\frac {0} {0}$.
$$\lim_{x \to 9} \frac {x-9} {\sqrt{x} – 3} = \frac{\lim_{x \to 9} (x-9) }{\lim_{x \to 9} {\sqrt{x} – 3}}$$
I mentioned that the limits do not exist because the function is not continuous at $ x=9$. Any idea on how to solve this question?
Best Answer
$$\frac{x-9}{\sqrt x-3}=\frac{(x-9)(\sqrt x+3)}{(x-9)}=\sqrt x+3\text{ if } x\ne 9$$
Now, if $x\to 9,x\ne 9$
So, $$\lim_{x\to 9}\frac{x-9}{\sqrt x-3}=\lim_{x\to 9}(\sqrt x+3)=\sqrt 9+3=6$$
Alternatively, let's consider $$\lim_{x\to a^2}\frac{x-a^2}{\sqrt x-a}$$
If we put $h=\sqrt x-a$ so that $h\to0$ as $x\to a^2$ and $x=(a+h)^2$
So, $$\lim_{x\to a^2}\frac{x-a^2}{\sqrt x-a}=\lim_{h\to0}\frac{(a+h)^2-a^2}h=2a \text{ as }h\to0\implies h\ne0$$
Here $a=3$
Also observe that $$\lim_{h\to0}\frac{(a+h)^2-a^2}h=\frac{d (x^2)}{dx}_{(\text{ at }x=a) }$$
by the Differentiation from First Principles or Definition via difference quotients.