4.7 | L’Hospitals Rule, Growth, and Dominance

    l’Hopital’s Rule
    Suppose that \(f\) and \(g\) are differentiable in an open interval \(I\) that contains \(a\), with the possible exception of \(a\) itself, and \(g'(x) \not = 0\) for all \(x\) in \(I\). If \(\displaystyle \lim_{x \rightarrow a} \frac{f(x)}{g(x)} \) is an indeterminate form of the type \(0/0\) or \(\infty/\infty\), then
    \lim_{x \rightarrow a} \frac{f(x)}{g(x)} = \lim_{x \rightarrow a} \frac{f'(x)}{g'(x)}
    provided that the limit on the right-hand side exists or is infinite.
    \lim_{x \rightarrow \infty} \frac{\ln(x^8-4)}{\ln(x) \cos(1/x)}

    Since the numerator and denominator both go to infinity, we have the indeterminate form \(\infty/\infty\).

    \lim_{x \rightarrow \infty} \frac{\ln(x^8-4)}{\ln(x) \cos(1/x)} & =
    \lim_{x \rightarrow \infty} \frac{\frac{d}{dx}[\ln(x^8-4)]}{\frac{d}{dx}[\ln(x) \cos(1/x)]} \\
    & = \lim_{x \rightarrow \infty} \frac{\frac{8x^7}{x^8-4}}{\frac{1}{x} \cos(1/x)+\frac{\ln x}{x^2} \sin(1/x)} \\
    & = \lim_{x \rightarrow \infty} \frac{8x^9}{(x^8-4)(x \cos(1/x)+ \ln x \sin(1/x))} \\
    & = \lim_{x \rightarrow \infty} \frac{8x^9}{(x^8-4)x \cos(1/x)} \\
    & = \lim_{x \rightarrow \infty} \frac{8x^9}{(x^8-4)x} \\
    & = \lim_{x \rightarrow \infty} \frac{8x^9}{x^9-4x} \\
    & = 8

    • in the first equality we applied l’Hospital’s Rule
    • in the second equality we evaluated the derivatives
    • in the third equality we made some algebraic simplifications
    • in the fourth equality we used the fact that
      \lim_{x \rightarrow \infty} \ln x \sin(1/x) = 0
    • in the fifth equality we used the fact that
      \lim_{x \rightarrow \infty} \cos(1/x) = 1
    • in the sixth equality we made algebraic simplifications
    • and for the last step, the limit at infinity of a rational function is the limit of the ratio of its leading terms

    An argument using end behavior is a faster alternative to the argument above.

    \frac{\ln(x^8-4)}{\ln(x) \cos(1/x)} \approx \frac{\ln(x^8)}{\ln(x) \cos(0)} = \frac{8\ln(x)}{\ln(x)} = 8
    \lim_{x \rightarrow \infty} \frac{\ln(x^8-4)}{\ln(x) \cos(1/x)} = \lim_{x \rightarrow \infty} 8 = 8.

    E 4.7 Exercises