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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 $x$.

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    $\begingroup$ Multiply and divide by $\sqrt{1+3x}+1$. $\endgroup$ – Siminore Sep 14 '14 at 15:35
  • $\begingroup$ 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.) $\endgroup$ – ys wong Sep 14 '14 at 15:38
  • $\begingroup$ 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 $\endgroup$ – user116160 Sep 14 '14 at 15:41
  • $\begingroup$ btw how do i phrase my limit like that? $\endgroup$ – user116160 Sep 14 '14 at 15:43
  • $\begingroup$ You can read up on it though. It will may things a lot easier. $\endgroup$ – ys wong Sep 14 '14 at 15:44
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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...

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

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

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

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