# L'Hopital's rule vs limits at infinity

I am gettind confused with regards to L'Hopitals rule and finding horizontal asymptotes as x -> infinity.

I have seen the method when we compare the denominator and numerator of a polynomial rational function and take teh vraiabe with the highest exponent and divide all elemenets of the eequation by this. From this we get one of three answers . a) teh answer is the ratio between the leading coefficients of the numerator and denominator 2) the answer is 0 and 3) the answer is none (no horizontal asymptote).

Based on having those above three options availabel to me , I am confused when I would then use L'Hopital. I have had it explained to use it when I get 0/0 or infinity/infinity (negative or positive) and a few other instances (product and difference).

• There are functions with horizontal asymptotes, but those functions do not contain rational functions. This idea with "dividing out the highest exponent" would no longer work. Take for example $y=xsin\frac{1}{x}$ and let $x$ go to infinity. – imranfat Apr 8 '15 at 22:00
• so x sin(1/x) woudl be a case of wher L'Hopital applies ? In the case where its rational functions though I dont understand how we know which to apply , L'Hopital or the "dividing out" method. – user3754366 Apr 8 '15 at 22:08
• Almost, The x upfront can be written as $\frac{1}{x}$ in a denominator. It is then a situation zero over zero on which the Rule applies. It can also be done without the Rule by setting $x=\frac{1}{t}$ You obtain a standard limit to which the answer is 1 – imranfat Apr 8 '15 at 22:10

Calculating the horizontal asymptote is essentially a measure of what happens to the function for large values of x in the domain.

Essentially you are finding the limit when f(x) approaches positive and negative infinity.

Therefore you don't use L'Hopitals Rule unless you get an indeterminate value when calculating the limit. These would include 0^infinity, infinity x infinity, infinity/0 etc.

• The indeterminate values are $1^\infty, 0^0, 0\times\infty, \frac{0}{0}, \frac{\infty}{\infty}$. $0^\infty$ is not indeterminate and would tend to $0$, $\infty\times\infty$ and $\frac{\infty}{0}$ tend to $\infty$. – Element118 Oct 11 '15 at 4:56