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## Homework Statement

Let [tex]r \in \mathbb{R}[/tex]. Show that [tex]\lim_{x \to +\infty} x^r / e^x = 0[/tex]

## Homework Equations

## The Attempt at a Solution

Intuitively, this is clear since exponential growth (i.e. denominator) is greater than linear growth (i.e. numerator).

If [tex]r \in \mathbb{N}[/tex] then it is easy. We just apply L'Hospital's Rule [tex]r[/tex] times and then we get that [tex]\lim_{x \to +\infty} x^r / \mathrm{e}^x = \lim_{x \to +\infty} rx^{r - 1} / \mathrm{e}^x =\ldots = \lim_{x \to +\infty} r! \cdot 1 / \mathrm{e}^x = 0 [/tex] since [tex]\lim_{x \to +\infty} \mathrm{e}^x = +\infty[/tex].

However, the issue that I have is when [tex]r \in \mathbb{R}[/tex]. The key step when [tex]r \in \mathbb{N}[/tex] was that I could keep taking derivatives until the [tex]x[/tex] term on the numerator becomes 1, and then all we're left is some constant on the numerator and something going to infinity on the denominator and hence the whole term tends to zero. This will clearly not work for when [tex]r \in \mathbb{R}[/tex]; i.e. if [tex]x^\sqrt{2}[/tex], then even if I take [tex]k \in \mathbb{N}[/tex] derivatives, I get something like [tex]\sqrt{2} (\sqrt{2} - 1) \ldots (\sqrt{2} - k) \cdot x^{\sqrt{2} - k}[/tex] and so the x term doesn't become 1, since [tex]\sqrt{2} \in \mathbb{R} / \mathbb{Q}[/tex] and [tex]k \in \mathbb{N}[/tex], it follows that [tex]\sqrt{2} - k \ne 0[/tex].

My attempt is the following but progress is limited. We write, [tex]x^r =\mathrm{e}^{r \ln x}[/tex], then it follows that [tex]x^r / e^x = \mathrm{e}^{r \ln x} / \mathrm{e}^x = \mathrm{e}^{r \ln x - x }[/tex]. Now, using the fact that [tex]\mathrm{e}^x[/tex] is continuous on [tex]\mathbb{R}[/tex], if we can show that [tex]\lim_{x \to +\infty} (r \ln x - x) = -\infty[/tex] then we are done.

To do this, we have to use an indeterminate form [tex]0 / 0 [/tex] or [tex]\infty / \infty[/tex] in preparation for L'Hospital's Rule. Note that if [tex]x = 0[/tex] then the result is trivially true, so WLOG, we can assume [tex]x \in \mathbb{R} - \{0\}[/tex]. Hence, we can write [tex]r \ln x - x = \frac{ \frac{r}{x} \ln x - 1} {\frac{1}{x}}[/tex]. Now, it is clear that on the denominator, [tex]\lim_{x \to +\infty} 1 / x = 0[/tex]. It remains to check check the hypothesis of L'Hospital's Rule and so we need to check that [tex]\lim_{x \to +\infty} (\frac{r}{x} \ln x - 1) = 0[/tex] holds..... this is exactly where I'm stuck.

Note that since by L'Hospital's Rule, [tex]\lim_{x \to +\infty} \frac{r}{x} \ln x = \lim_{x \to +\infty} r \cdot \frac{1 / x}{1} = r \cdot 0 = 0[/tex]. And it follows that [tex]\lim_{x \to +\infty} (\frac{r}{x} \ln x - 1) = 0 - 1 = -1[/tex]. Thus, we have an indeterminate form [tex]-1 / 0[/tex] and hence, this fails the hypothesis of L'Hospital's Rule! And note that I'm very hesitant to conclude that [tex]-1 / 0 \approx -\infty[/tex] since I know this is not true...

Any help is appreciated!

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