$\begingroup$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$3 Answers
$\begingroup$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$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$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.
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