<s>[INST] Solving $\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$ without L'Hopital Without L'Hopital:
$$\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}$$
Rationalize:
$$\frac{x}{\sqrt{1-3x}-1}\cdot \frac{\sqrt{1-3x}+1}{\sqrt{1-3x}+1}$$
$$\frac{x\cdot(\sqrt{1-3x}+1)}{(1-3x)-1}$$
This will still yield $\frac{0}{0}$. Maybe I should now try variable substitution to eliminate the root. Let
$$w^2 = 1-3x$$
So
$$w\to 1$$
Therefore
$$\lim_{w\to1}\frac{x\cdot(\sqrt{w^2}+1)}{(1-3x)-1} = \lim_{w\to1}\frac{x\cdot(w+1)}{(1-3x)-1}$$
We find that
$$x = \frac{1-w^2}{3}$$
So
$$\lim_{w\to1}\frac{\left( \frac{1-w^2}{3}\right)\cdot(w+1)}{\left(1-3\left( \frac{1-w^2}{3}\right)\right)-1}$$
$$\lim_{w\to1}\frac{\left( \frac{1-w^2}{3}\right)\cdot(w+1)}{\left(1-3+3w^2\right)-1}$$
This will still evaluate to $\frac{0}{0}$.
I just usted the two methods that I always use to calculate limits. Rationalize/factorize and variable substitution. I'm not supposed to use L'Hopital.
What did I do wrong, and how should I have done it?
 [/INST] Let $\sqrt{1-3x}=1+h$. We know that when $x=0$ the LHS $=1 \Rightarrow h=0$ 
Then $1-3x=1+2h+h^2$
$-3x=2h+h^2$
$x=\frac{2h+h^2}{-3}$
The expression we want is $\frac x{\sqrt{1-3x}-1}=\frac {\frac{2h+h^2}{-3}}{1+h-1}=\frac{2+h}{-3}$.
As $h$ tends to $0$ this tends to $-\frac23$ 
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