7
$\begingroup$

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?

$\endgroup$
7
$\begingroup$

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 $y=2x+\dfrac{1}{4}$

$\endgroup$
2
  • $\begingroup$ Just a question I'm having, shouldn't a = lim (f(x)-b)/x? $\endgroup$ – Nima G Oct 7 '12 at 16:24
  • 3
    $\begingroup$ @Nima They are the same thing since $\lim b/x\to 0$ $\endgroup$ – user39572 Oct 7 '12 at 16:29
4
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Very interesting, thanks Lubin. $\endgroup$ – Nima G Oct 2 '12 at 18:44
0
$\begingroup$

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 $$f(x) - (2x + 1/4) = \frac{95/16}{f(x) + (2x + 1/4)}.$$

But $f(x) + (2x + 1/4) \rightarrow \infty$ as $x \rightarrow \infty$. This implies that $$f(x) - (2x + 1/4) \rightarrow 0$$ as $x \rightarrow \infty$.

Remark: Likewise $$f(x) + (2x + 1/4) = \frac{95/16}{f(x) - (2x + 1/4) }$$ and hence as $x \rightarrow -\infty, f(x) - (2x + 1/4) \rightarrow \infty$ and hence $$\frac {95/16}{f(x) - (2x + 1/4)} \rightarrow 0.$$ This implies that $f(x) + (2x + 1/4) \rightarrow 0$ and thus $y = -(2x + 1/4)$ is another asymptote.

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.