Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 410: 59

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Here, we have $\ln f(x)=x \ln x$ But $\lim\limits_{x \to 0} f(0)=\dfrac{\infty}{\infty}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ $e^{\lim\limits_{x \to 0} \dfrac{1/x}{-1/x^2}}=e^{\lim\limits_{x \to 0} (-x)}$ and $e^{0}=1$