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I'm looking for a way of finding an oblique asymptote of (on infinity):

\begin{equation} \sqrt{1 + x^2 + \sqrt{(1 + x^2)^2 - 2 x^2 \cos^2{\theta}}} \end{equation}

I know that the asymptote is $\sqrt{2}x$.

I'm trying to find it simply by finding limit on the infinity:

\begin{align} \lim\limits_{x\to\infty}\sqrt{1 + x^2 + \sqrt{(1 + x^2)^2 - 2 x^2 \cos^2{\theta}}} & = |\text{Omitting limit sign for a while}| = \\ = |\text{Take }\theta\text{ be equal to it max value}| &= \sqrt{1 + x^2 + \sqrt{1 + 2 x^2 + x^4 - 2 x^2}} = \\ = |\text{Omitting 1}| & = \sqrt{x^2 + \sqrt{x^4}} = \\ = |\text{Simplifying}| & = \sqrt{2} x \end{align}

It this way of finding this asymptote is correct? I'm mostly confused with the step when I'm omit the $1$ comparing to $x^n$.

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1 Answer 1

up vote 1 down vote accepted

Basically, yes. The reason the $\cos$ term doesn't matter is not that we may substitute with its maximum value (this is a limit, not an inequality), but that the degree of the monomial it's attached to is found from $(x^2)^{1/4} = x^{1/2}$, and is smaller than the maximum degree involved, which (as you found) is 1. The reason that 1 vanishes compared to $x^n$ for positive $n$ is that the degree of $x^n$ is, well, $n$, and the degree of $1 = x^0$ is $0<n$.

The $\sqrt 2$ coefficient comes from noticing there are two terms of maximum degree (the $x^2$ under the square root, and the term of order 4 in $(1+x^2)^2$ under the fourth root), allowing all other terms to vanish, and computing the result as though we set $x=1$. The logic behind this is the same as multiplying-and-dividing by $x$ to the maximum-degree-power, evaluating the limit, and seeing what vanishes. To see this in action, we have that the coefficient is found by $$ \sqrt{1+x^2+\sqrt{(1+x^2)^2-2x^2\cos\theta}} \to \sqrt{x^2+\sqrt{x^4}} \to \sqrt{1+\sqrt{1}} = \sqrt 2~~. $$

We know that the power of $x$ will be 1 because that is the maximum degree present in the expression. Note that this works for any non-rational expression involving monomial powers (for a rational expression, one must compute separately for top and bottom). All oblique asymptotes found this way will obey a power law; i.e. $f\sim ax^n$

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