# How to find the oblique asymptote of root of a function?

In a test example I'm solving, the question asks to find the oblique asymptote of the following function:

$f(x) = \sqrt{4x^2+x+6}$

$x$ at $+\infty$

We have only learned how to do so with rational functions. Is there any general way of finding the oblique asymptote that works with any kind of function? Perhaps using limits?

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Yes. If $f$ has an oblique asymptote (call it $y=ax+b$), you will have: $$a=\lim_{x\to\pm\infty}\frac{f(x)}{x}$$

$$b=\lim_{x\to\pm\infty} f(x)-ax$$

In your example, $\displaystyle\lim_{x\to+\infty}\frac{\sqrt{4x^2+x+6}}{x}=2$ and $\displaystyle\lim_{x\to+\infty}\sqrt{4x^2+x+6}-2x=\frac{1}{4}$

The asymptote as $x\to+\infty$ is therefore $y=2x+\dfrac{1}{4}$

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Just a question I'm having, shouldn't a = lim (f(x)-b)/x? –  Nima Oct 7 '12 at 16:24
@Nima They are the same thing since $\lim b/x\to 0$ –  user39572 Oct 7 '12 at 16:29

The answer of @Julien is perfect, but here’s another outlook. Take your function, and factor out $4x^2$ from the radicand, getting $2x\sqrt{1+1/(4x) + 3/(2x^2)}=2x(1+\frac{1}{4}x^{-1}+\frac{3}{2}x^{-2})^{1/2}$. For the (positive) asymptote, you’re interested in cases where $x^{-1}$ is tiny, so you can approximate the radical very well with the Taylor expansion $(1+A)^{1/2}=1+\frac{1}{2}A-\frac{1}{8}A^2+\cdots$. Setting $A=\frac{1}{4}x^{-1}+\frac{3}{2}x^{-2}$ and looking only at the constant and the $x^{-1}$-term, you get $2x(1+\frac{1}{8}x^{-1}+\cdots)$, the same result that @Julien announced.

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Very interesting, thanks Lubin. –  Nima Oct 2 '12 at 18:44

By completing the square we get f(x) = sqrt( (2x + 1/4)^2 + 95/16).

This means (after squaring both sides and taking (2x + 1/4)^2 to the left hand side and factoring) that ( f(x) - (2x + 1/4) ) ( f(x) + (2x + 1/4) ) = 95/16

And hence f(x) - (2x + 1/4) = (95/16) / ( f(x) + (2x + 1/4) )

But f(x) + (2x + 1/4) -----> infinity as x ----> infinity. This implies that

                  f(x) - (2x + 1/4)  ----> 0  as x ---> infinity

Remark: Likewise    f(x) + (2x + 1/4)  =  (95/16) / ( f(x) - (2x + 1/4) )


and hence as x ----> -infinity , ( f(x) - (2x + 1/4) ) ---> infinity and hence

(95/16) / ( f(x) - (2x + 1/4) ) ---> 0 . This impolies that f(x) + (2x + 1/4) ---> 0

and thus y = -(2x + 1/4) is another asymptote.

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No, $f(x) - (2x+1/4) \to 0$, not $\infty$. $-(2x+1/4)$ is not another asymptote. –  Antonio Vargas Nov 19 '13 at 1:46

When x ---> minus infinity f(x) and - (2x +1/4) ---> infinity and hence their sum i.e.

f(x) - (2x + 1/4) ---> infinity and that's why f(x) + (2x + 1/4) ---> 0 and hence

yes, y = - (2x + 1/4) is another asymptote.

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