$$y=\dfrac{(\sin x)^3}{x^2}$$ My professor explained how to do this problem by using limits. I am a little confused as to why he used limits because in the past we were told to look at the coefficients of the biggest degrees. I have done some searching online and found that if you take the limit of a function as it approaches $\infty$ or $-\infty$, you would get the horizontal asymptotes. I have also came across someone saying that the coefficients do not matter and only the degrees matter. Is this true for all cases? In the past I was told to use the ratio of the coefficients of the largest degree and that would be the horizontal asymptote. Is this not the case when you are using limits to find the horizontal asymptotes? If not, could someone give me an example when the coeffcients are used?


You can look at the degrees if the numerator and denominator are both polynomials. In general, it's not so simple; in general there isn't even anything you can identify as a "degree". Here, the numerator is not a polynomial, and it's not so simple.

If you try to pretend that $(\sin x)^3$ is a third-degree polynomial, you will get the wrong answer.

You can use the coefficients of the highest-degree terms of the numerator and denominator when the numerator and denominator are polymomials. For example,


has a horizontal asymptote at $y=4$, and


has one at $y=0$.

  • $\begingroup$ I know that the $sin$ function is bounded between $-1$ and $1$ so the limit of the function approaches $0$. I have a couple of questions that I asked and I was wondering if you could explain it to me. $\endgroup$ – Kot Feb 28 '13 at 3:16
  • $\begingroup$ Ahh I see now! Thank you for the explanation :). $\endgroup$ – Kot Feb 28 '13 at 3:22

Consider the degrees of the numerator and denominator when both are polynomials. If they have the same degree, we only need to consider the coefficients of the highest powers; however, if their degrees differ, then either the numerator has a higher power or a lower power. In the former case, the limit tends to $\infty$ or $-\infty$, depending on the sign of its coefficient ($+\infty$ if the coefficient is positive, and $-\infty$ if it is a negative coefficient). Consider, for example, $$\lim_{x\to\infty}\frac{x^2+1}{x+1}.$$ The numerator has a higher power, and the numerator and denominator are positive, hence the limit is $+\infty$.

In the latter case, the limit always goes to zero, as in the example $$\lim_{x\to\infty}\frac{x+1}{x^2+1}.$$

Note that if the numerator has a higher power by only $1$ degree, we can use long division of polynomials to actually calculate the oblique asymptote. Considering the first example, we can calculate $$\frac{x^2+1}{x+1}=x-1+\frac{2}{x+1}.$$ As $x$ gets big, the last term goes to zero, thus the oblique asymptote is $$x-1.$$

In cases where they're not both polynomials, we usually have to rely on some bounds. In your example, we know $-1\leq\sin(x)\leq 1$, and the limits become easy to calculate. That is, we have $$\lim_{x\to\infty}\frac{-1}{x^3}=0\leq\lim_{x\to\infty}\frac{\left(\sin x\right)^3}{x^3}\leq0=\lim_{x\to\infty}\frac{1}{x^3}$$

  • $\begingroup$ Is the $-1\leq\sin(x)\leq 1$ true for all trigonometric functions and are there any other types of bounds I should be aware of? I have an exam this Friday and I want to be sure I know this. $\endgroup$ – Kot Feb 28 '13 at 3:37
  • $\begingroup$ @StevenN: It is true only for sine and cosine; the tangent function has its range as $\Bbb R$, so there cannot be bounds for it, and secant and cosecant have their ranges as $(-\infty,-1]\cup[1,\infty)$. $\endgroup$ – Clayton Feb 28 '13 at 3:41
  • $\begingroup$ I see, so if the $sin$ were replaced with $tan$, $csc$ and $sec$ in the problem above, the limits would also be $0$? Or would the power in the numerator play a role in this? Also, if the $sin$ and $x^2$ were switched, it would shoot up to infinity and have no horizontal asymptote right? $\endgroup$ – Kot Feb 28 '13 at 3:44
  • $\begingroup$ @StevenN: If $\sin$ were replaced by any of those, the problem changes drastically, as we no longer have the trig function bounded. For the second part, if we switched places for $x^3$ and $\sin(x)^3$, we would get that the limit diverges since no matter how you look at it, the denominator bounces back and forth between $1$ and $-1$, thus our limit has two subsequences, one 'converging' to $\infty$ and one converging to $-\infty$. $\endgroup$ – Clayton Feb 28 '13 at 4:05
  • $\begingroup$ I guess I am getting a bit ahead of myself. I hope that the other trigonometric functions won't be on my test. $\endgroup$ – Kot Feb 28 '13 at 4:17

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