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How to find the Horizontal asymptote for this function:

$f(x)= \sqrt {9x^2+2x}-3x$

I have tried to find the lim where x goes to infinity and negative infinity then what ?

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  • $\begingroup$ This function hasn't horizontal asymptote... $\endgroup$ – alexjo Dec 7 '13 at 13:41
  • $\begingroup$ Did you mean $\sqrt{9x^2+2x}-3x$? $\endgroup$ – robjohn Dec 7 '13 at 14:04
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Assuming you meant $f(x)=\sqrt{9x^2+2x}-3x$, we can write $$ \begin{align} \sqrt{9x^2+2x}-3x &=\left(\sqrt{9x^2+2x}-3x\right)\frac{\sqrt{9x^2+2x}+3x}{\sqrt{9x^2+2x}+3x}\\ &=\frac{2x}{\sqrt{9x^2+2x}+3x}\\ &=\frac{2}{\sqrt{9+2/x}+3}\\ &\to\frac13 \end{align} $$ The technique of multiplying $\left(\sqrt{a}-\sqrt{b}\right)$ by $\frac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}}$ to get $\frac{a-b}{\sqrt{a}+\sqrt{b}}$ is very useful in a number of problems.

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  • $\begingroup$ would the downvoter care to comment? $\endgroup$ – robjohn Dec 8 '13 at 23:09

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