Infinity over Infinity

This is another common use of l'Hôpital's Rule. If you are finding a limit of a fraction, where the limits of both the numerator and the denominator are infinite, then l'Hôpital's Rule says that the limit of the fraction is the same as the limit of the fraction of the derivatives. For example,

the limit as x goes to infinity of x/(exp(x)) *= the limit as x goes to infinity of 1/(exp(x)) = 0.

Note that both x and e^x approach infinity as x approaches infinity, so we can use l'Hôpital's Rule. Also, the derivative of x is 1, and the derivative of e^x is (still) e^x.

Here is another example.

the limit as x goes to infinity of (x^2-4)/(x^2-3x+2) *= the limit as x goes to infinity of (2x)/(2x-3)

Note 2x is the derivative of x^2 - 4, and 2x - 3 is the derivative of x^2 - 3x + 2. Now, the limits of both 2x and 2x - 3 are still infinite, so we can use l'Hôpital's Rule again:

the limit as x goes to infinity of (2x)/(2x-3) *= the limit as x goes to infinity of 2/2 = 1.

Here is a more complicated example:

the limit as x goes to zero from the right of ln(x)/cot(x) *= the limit as x goes to 0 from the right of (1/x)/(-csc^2(x)) = the limit as x goes to 0 from the right of (-sin^2(x))/x.

Here we use the fact that csc(x) = 1/sin(x) to simplify the fraction. Note that for the last fraction, the limit of the numerator and denominator are both zero, which is another case where we can apply l'Hôpital's Rule.

the limit as x goes to 0 from the right of (-sin^2(x))/x *= the limit as x goes to 0 from the right of (-2*sin(x)*cos(x))/1 = (-2*0*1)/1 = 0

Copyright © 1996 Department of Mathematics, Oregon State University

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