# Evaluation of a limit using L'Hopital's Rule

I would like to know if my understanding of the following is correct. This has been tripping me up for a long time now.

Compute $\lim_{x\rightarrow \infty}x^{1-\beta}$.

This is part of a homework problem regarding the expectation of the Pareto distribution. It says everywhere that the expectation is for $\beta>1$ only, but what about $B=1$? Then doesn't the expectation change significantly because as you all have said, the limit at $\beta = 0$ evaluates to 1?

Could someone please clarify my doubt? Thanks.

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For $\beta = 1$, you have $\lim\limits_{x\to\infty} x^0 = \lim\limits_{x\to\infty} 1 = 1$. – Daniel Fischer Oct 3 '13 at 3:12
L'Hospital's Rule is not the right tool. – André Nicolas Oct 3 '13 at 3:15
I have edited my question significantly to ask a more pertinent question. – Anonymous Oct 3 '13 at 3:23

Let $$f = \frac{x}{x^{\beta}}$$

If $\beta = 1$, then $\lim f = \lim 1 = 1$

If $\beta > 1$, then $\beta - 1 > 0 \implies \lim f = \lim \frac{1}{x^{\beta - 1}} = 0$

If $\beta < 1$, then $1 - \beta > 0 \implies \lim f = \lim x^{1 - \beta} = \infty$

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Not quite: If $\beta = 1$, then $x^{1 - \beta} = x^0 = 1$ for all $x > 0$; hence, the limit is

$$\lim_{x \to \infty} x^{1 - \beta} = \lim_{x \to \infty} 1 = 1$$

You are correct for the other two, though.

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