Find limit when $x\to0$ $$\lim_{x\to0} \frac{\sqrt{2x + 9} - 3}x$$
I tried to solve this but squaring both values but I cant get rid of the square root.
How to solve this?
 A: $$\frac{\sqrt{2x+9}-3}x=\frac{2x+9-9}{x(\sqrt{2x+9}+3)}=\frac2{\sqrt{2x+9}+3}\text{ if }x\ne0$$
Now we know if $x\to0,x\ne0$
Can you take it from here?
A: If you've gotten to differentiation yet, there's an "obvious" thing to do here.
The problem is that both $\sqrt{2x+9}$ and $3$ converge to the same value as $x \to 0$, and subtracting the two cancels out the most significant parts of their values.
So, we need to figure out the next most significant parts of their values: differential approximation (or more generally Taylor series) is an excellent way to do this.
Recall that differential approximation says that, e.g.,
$$ f(x) = f(0) + f'(0) x + x r(x) $$
where $r$ is a function that satisfies
$$ \lim_{x \to 0} r(x) = 0 $$
If you let $f(x) = \sqrt{2x+9}$, you can use differential approximation to rewrite your limit to make it simpler.
A: The dominating element under the square root is the 9. If you factorize accordingly,
$$
\frac{3\left(\sqrt{1+2x/9}-1\right)}{x} \sim \frac{3}{x}\times\frac{1}{2}\frac{2x}{9} = \frac{1}{3}
$$
