Computing limit of $f(x) = x^{(\frac{1}{x} - 1)}$

Question

Let $f(x) = x^{(\frac{1}{x} - 1)}$ . Find $\lim_{x \to 0^{+}} f(x)$ if it exists .

My try : $f(x) = x^{(\frac{1}{x} - 1)} = e^{\frac{(1-x) \ln x}{x}}$ . I'm not allowed to use L'Hôpital's rule but intuitively seems the answer is zero .

Answer 1

This is not indeterminate when $x\to 0^+$:

• $\ln x\to -\infty$,
• $1-x\to 1$, so $(1-x)\ln x\to-\infty$
• the denominator tends to $0^+$ so the exponent tends to $-\infty$, and the exponential indeed tends to $0$.

Answer 2

$$\frac{(1-x)\log{x}}{x}=\frac{1-x}{x}\log{x}=(\frac{1}{x}-1)\log{x}$$

$$\lim_{x\to 0^{+}}{(\frac{1}{x}-1)}=+\infty \text{ and } \lim_{x\to 0^{+}}{\log{x}}=-\infty$$

$$\lim_{x\to 0^{+}}{(\frac{1}{x}-1)\log{x}}=-\infty$$

$$\lim_{x\to 0^{+}}{x^{\frac{1}{x}-1}}=\lim_{x\to 0^{+}}{e^{\frac{(1-x)\log{x}}{x}}}=0$$

Answer 3

Hint: remember $\infty \times - \infty= -\infty.$Take $e$ and $\ln$. Also, since $e$ is a continuous function so it behaves nicely with limit. I hope it should be clear now.