lim
One way to think of this problem is in terms of the relative growth rates between the numerator and denominator. I know that x grows asymptotically faster than (\ln x)^3 according to WolframAlpha. How can I prove this?
Answer
\lim_{x \to \infty} \frac{x}{(\ln x)^3}=\quad\quad (\frac{\infty}{\infty}\textrm {form, using L'Hospital rule})\\ =\lim_{x \to \infty}\frac{x}{3(\ln x)^2}\quad\quad (\frac{\infty}{\infty}\textrm {form, using L'Hospital rule})\\=\lim_{x \to \infty}\frac{x}{6(\ln x)}\quad\quad (\frac{\infty}{\infty}\textrm {form, using L'Hospital rule})\\ =\lim_{x \to \infty}\frac{x}{6}=\infty
No comments:
Post a Comment