# How to evaluate $\lim_{x\to 0}x/( \sqrt{1+3x}-1)$ using limit laws

I have been looking over the limit laws and watching videos but I can't find a similar problem. The question asks me to use limit laws to evaluate the limit

$$\lim_{x\to 0}\frac{x}{\sqrt{1+3x}-1}$$

I tried rationalizing the denominator but multiplied the top out instead of cancelling the $x$.

• Multiply and divide by $\sqrt{1+3x}+1$. – Siminore Sep 14 '14 at 15:35
• Use Lhospital rule. Differentiate the top and bottom and then take limit again. The easiest way is just to use L hospital rule.(not just for this limit question but for many other limit question as well.) – ys wong Sep 14 '14 at 15:38
• i just started calc a week ago im in and in the hospital rule isnt in the chapter section so i assume im not supposed to know it yet – user116160 Sep 14 '14 at 15:41
• btw how do i phrase my limit like that? – user116160 Sep 14 '14 at 15:43
• You can read up on it though. It will may things a lot easier. – ys wong Sep 14 '14 at 15:44

If you Multiply and divide by $\sqrt{1+3x}+1$ you get $$\frac{x (\sqrt{1+3x}+1)}{1+3x-1} = \frac{1}{3} (\sqrt{1+3x}+1),$$ which tends to...

$$\frac{x}{\sqrt{1+3x}-1}=\frac{x\left(\sqrt{1+3x}+1\right)}{\left(\sqrt{1+3x}-1\right)\left(\sqrt{1+3x}+1\right)}=\frac{\left(\sqrt{1+3x}+1\right)}{3x}=\frac{\sqrt{1+3x}+1}{3}$$

Then$$\lim_{x\to 0}\frac{x}{\sqrt{1+3x}-1}=\lim_{x\to 0}\frac{\sqrt{1+3x}+1}{3}=\frac{2}{3}$$

$$\frac x{\sqrt{1+3x}-1}\cdot\frac{\sqrt{3x+1}+1}{\sqrt{3x+1}+1}=\frac{\color{red}x\left(\sqrt{3x+1}+1\right)}{3\color{red}x}=\ldots$$