# Lim_{x to infty} x^r / e^x = 0, where r is real

1. Sep 21, 2009

### fmam3

1. The problem statement, all variables and given/known data
Let $$r \in \mathbb{R}$$. Show that $$\lim_{x \to +\infty} x^r / e^x = 0$$

2. Relevant equations

3. The attempt at a solution
Intuitively, this is clear since exponential growth (i.e. denominator) is greater than linear growth (i.e. numerator).

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

However, the issue that I have is when $$r \in \mathbb{R}$$. The key step when $$r \in \mathbb{N}$$ was that I could keep taking derivatives until the $$x$$ 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 $$r \in \mathbb{R}$$; i.e. if $$x^\sqrt{2}$$, then even if I take $$k \in \mathbb{N}$$ derivatives, I get something like $$\sqrt{2} (\sqrt{2} - 1) \ldots (\sqrt{2} - k) \cdot x^{\sqrt{2} - k}$$ and so the x term doesn't become 1, since $$\sqrt{2} \in \mathbb{R} / \mathbb{Q}$$ and $$k \in \mathbb{N}$$, it follows that $$\sqrt{2} - k \ne 0$$.

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

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

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

Any help is appreciated!

Last edited: Sep 21, 2009
2. Sep 21, 2009

### zcd

Why not just apply l'hopital's rule from the start?
If we assume r to be a positive integer:
$$\lim_{x\to\infty} \frac{x^{r}}{e^{x}}=\lim_{x\to\infty} \frac{rx^{r-1}}{e^{x}}=\lim_{x\to\infty} \frac{r(r-1)x^{r-2}}{e^{x}}=\lim_{x\to\infty} \frac{r(r-1)(r-2)...(2)(1)x^{0}}{e^{x}}=\lim_{x\to\infty} \frac{r!}{e^{x}}=0$$
To extend this to positive numbers, you can take replace $$r!=\Gamma(r+1)$$. If it's negative, then it'll go to zero already.

3. Sep 21, 2009

### fmam3

Thanks for the reply! Actually, as I'd noted, if $$r \in \mathbb{N}$$ then I already know how to solve it --- in fact, using the same method you used. But my problem is that it is given that $$r \in \mathbb{R}$$; again, as I'd noted, if, say, $$r = \sqrt{2}$$, then no matter how many derivatives you take on $$x^\sqrt{2}$$, we will never get something like $$x^0 = 1$$.

Any further help is appreciated!

4. Sep 21, 2009

### zcd

As already said, replace $$r!=\Gamma(r+1)$$. The gamma function is an extension of factorials from integers, and yields a positive real number for values >0.

5. Sep 21, 2009

### fmam3

Is it possible to solve the problem without using the Gamma function? Since it is not introduced at this point....

6. Sep 21, 2009

### zcd

Let x=lnt, then $$\lim_{t\to\infty} r\frac{\ln(t)}{t}=\lim_{t\to\infty} r\frac{1}{t}=0$$

7. Sep 21, 2009

### fmam3

What does "let x = Int" mean?

8. Sep 21, 2009

### zcd

Define t as a variable which depends on x. The exact relationship is $$t=e^{x}$$. Since t and ex are equal, you can substitute it back into the limit.

9. Sep 21, 2009

### fmam3

Actually, on another thought of this --- how does, even with the use of the Gamma function, allow you to show this, when $$r \in \mathbb{R}$$. That is, the key part of your argument, when you'd assumed r is a positive integer is that this holds: $$\lim_{x\to\infty} \frac{x^{r}}{e^{x}}=\lim_{x\to\infty} \frac{rx^{r-1}}{e^{x}}=\lim_{x\to\infty} \frac{r(r-1)x^{r-2}}{e^{x}}=\lim_{x\to\infty} \frac{r(r-1)(r-2)...(2)(1)x^{0}}{e^{x}}=\lim_{x\to\infty} \frac{r!}{e^{x}}=0$$, and the key is that ultimately, you get to a term where $$x^0 = 1$$.

Again, if $$r = \sqrt{2}$$, what is the addictive inverse in the natural numbers, say k, such that $$\sqrt{2} - k = 0$$ for $$k \in \mathbb{N}$$?

10. Sep 21, 2009

### zcd

I admit that I also find it questionable to just plug in the gamma function after I've declared r to be a positive integer, but if r is a positive noninteger then after enough derivatives it will pass zero and go into the negatives. When r is negative the limit will approach 0.

11. Sep 21, 2009

### fmam3

Again, I'm not disputing your argument (in fact, I agree with 120%) that it is true. With r positive or negative integer, it is trivially true.

The tough part of this question is dealing with cases like $$r = \sqrt{2}$$.

12. Sep 21, 2009

### fmam3

Sorry but I'm actually finding this hard to understand. Yes, if you set $$t = e^x$$, then as $$x \to +\infty$$, $$t = t(x) = e^x \to +\infty$$. But that does not mean $$t \to \infty$$ implies $$x \to \infty$$.

But the way you had applied L'Hospital's Theorem is incorrect. Suppose $$t = e^x$$, or $$x = \ln t$$, then we have that
$$\lim_{x \to +\infty} x^r / e^x = \lim_{x \to +\infty} r \ln x / e^x = \lim_{x \to +\infty} \frac{r \ln( \ln(t(x)) )}{ t(x)}$$. Then even if you apply L'Hospital's Rule, you need to take derivatives with respect to x, and not t, since t is a function of x....

13. Sep 21, 2009

### slider142

You do not need to use anything about the gamma function. Simply extend your integer argument using the squeeze theorem.

14. Sep 21, 2009

### fmam3

Thanks! That was the tip I'd needed to push me into the right direction.

The proof is as follows. By the Archimedian Property, for $$\forall r \in \mathbb{R}$$, $$\exists n \in \mathbb{N}$$ such that $$n \geq r$$. Then consider the limit value $$\lim_{x \to +\infty} x^n / e^x$$. This is an indeterminate form $$\infty / \infty$$ and hence we can apply L'Hospital's Rule. In fact, by applying L'Hospital's Rule n times, we have that $$\lim_{x \to +\infty} x^n / e^x = \lim_{x \to +\infty}nx^{n-1} / e^x = \ldots = \lim_{x \to +\infty} n!x^0 / e^x = n! / e^x = 0$$. This implies that $$\forall \varepsilon > 0, \exists \alpha > 0$$ such that $$\forall x > \alpha$$, we have $$| x^n / e^x - 0 | < \varepsilon$$. Then, it follows that for $$x > \max\{ \alpha, 1\}$$, we have that $$|x|^r \leq |x|^n$$, which implies $$0 < |x^r / e^x| \leq |x^n / e^x| < \varepsilon$$. Thus, by the Squeeze Theorem (or just evident by the expression above), it follows that $$\lim_{x \to +\infty} x^n / e^x = 0$$ as desired.

Thanks again!