Thursday, 5 September 2013

Limit limxtoinftyfrac(ln(x))kx




The limit is lim \forall k \in N, k \ge 1



For k = 1 or 2, the limit is fairly easy to solve using the Stolz–Cesàro theorem. Intuitively, because ln(x) grows so much slower than x, the answer should be 0. But how can we mathematically solve the limit for any positive integer k?


Answer



\lim_{x\to\infty}\frac{\ln^kx}{x}= \left(\lim_{x\to\infty}\frac{k\ln x^{\frac1{k}}}{x^{\frac1{k}}} \right)^k = k^k \left( \lim_{y\to\infty} \frac{\ln y}{y} \right)^k= 0


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