# Chapter 4 - Section 4.5 - Indeterminate Forms and L'Hôpital's Rule - Exercises - Page 248: 55

$\dfrac{1}{e}$

#### Work Step by Step

L'Hospital's rule states that $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{A'(x)}{B'(x)}$ Here, $\ln f(x)=(\ln x) ^{-1/\ln x} \implies\ln f(x)= \dfrac{-\ln x}{\ln x}$ or, $\ln f(x)=\frac{-(\ln x)}{\ln x} \implies f(x)=-1$ Now, $e^{\lim\limits_{x \to 0^{+}} (-1)}=e^{-1}=\dfrac{1}{e}$

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