Attempting to answer this question, I thought to evaluate the limit by taking the logarithm and then using L'Hopital's rule:

$$\begin{align} L&=\lim_{x\to\infty}\dfrac{e^x}{x^{x-1}}\\ \ln{L}&=\lim_{x\to\infty}\frac{x}{(x-1)\ln(x)} \\ \ln{L}&=\lim_{x\to\infty}\frac{1}{(x-1)\frac{1}{x}+\ln{x}} \\ \ln{L}&=\lim_{x\to\infty}\frac{1}{1-\frac{1}{x}+\ln{x}} \\ \ln{L}&=0 \\ e^{\ln{L}} &=e^0 \\ L&=1 \end{align}$$

But Wolfram Alpha says that the limit is $0$. Where am I going wrong?

  • 1
    $\begingroup$ $\ln(a/b)\ne \ln a/\ln b$. $\endgroup$ – David Mitra Jul 25 '13 at 16:00
  • $\begingroup$ See here for the techniques. $\endgroup$ – Mhenni Benghorbal Jul 25 '13 at 16:04

$$\ln(e^x/x^{x-1}) =x-(x-1)\ln x\neq \frac{x}{(x-1)\ln x}$$


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