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Why does $x^{(1/\ln(x))} = e$, espacially for a limit as it reaches infinity. I figured it would just be $0$ since $\frac{1}{\ln(\infty)}$ should equal $0$.

I don't get the concept.

rae306
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user175271
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    Yes, $1/\ln(x)$ goes to zero, but $x$ goes to infinity, so your looking at a $\infty^0$-limit. Those can go to more or less anything. In this case, it goes to $e$. – Arthur Sep 11 '14 at 10:33
  • Take the natural log of both sides and insight is not far off. – Tpofofn Sep 11 '14 at 10:45

3 Answers3

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If $x>0$ and $x\neq 1$ then:

$$x^{\frac{1}{\ln x}}=x^{\frac{\ln e}{\ln x}}=x^{\log_{x}e}=e$$

Practicized is the general rule: $$\log_{g}a=\frac{\log_{h}a}{\log_{h}g}$$ for any $h$ with $h>0$ and $h\neq1$.

drhab
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$x^{1/\ln x}=(\exp(\ln x))^{1/\ln x}=\exp(\ln x/\ln x)=\mathrm{e},\ x\ne0$

g.kov
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Another approach assuming you don't know the result of $x^{\frac{1}{\ln(x)}}$.

$$x^{\frac{1}{\ln(x)}}=y$$

$$x^{\frac{\ln(e)}{\ln(x)}}=y$$

$$\ln(x)^{\frac{\ln(e)}{\ln(x)}}=\ln(y)$$

$$\frac{\ln(e)}{\ln(x)}\ln(x)=\ln(y)$$

$$\ln(y)=\ln(e)$$

$$y=e$$