$$y=\dfrac{(\sin x)^3}{x^2}$$
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?
[Math] Concept of using limits to find horizontal asymptotes.
calculuslimits
Best Answer
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,
$$\frac{12x^2+7x+17}{3x^2-8}$$
has a horizontal asymptote at $y=4$, and
$$\frac{7x+17}{3x^2-8}$$
has one at $y=0$.