<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] In the beginning of your post, you have
$$\frac{x}{\sqrt{1-3x}-1}\cdot \frac{\sqrt{1-3x}+1}{\sqrt{1-3x}+1}=\frac{x\cdot(\sqrt{1-3x}+1)}{(1-3x)-1}=\frac{\sqrt{1-3x}+1}{-3}$$
for all $x\leq 1/3$.
Hence
$$\lim_{x\to0}\frac{x}{\sqrt{1-3x}-1}=\lim_{x\to0}\frac{\sqrt{1-3x}+1}{-3}.$$
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