Evaluating a limit that requires x-iterations of applying L'Hopital's Rule.

Question

So I need to evaluate this limit: $$\lim_{h\to\infty} \frac{h^y}{e^h}, \space\space\space y>0$$

If you apply l'Hopital's Rule you get: $$\lim_{h\to\infty} \frac{yh^{y-1}}{e^h}, \space\space\space y>0$$

I guess, you could say that l'Hopital's rule should be applied $y$ times until the limit becomes trivial, but is there a more elegant way or working this out?

BTW, $y$ is finite and positive but not known.

EDIT, as suggested:

$${\lim_{h\to\infty} \frac{h}{e^\frac{h}{y}}}^y$$

Then using l'Hopital's Rule:

$${\lim_{h\to\infty} \frac{y}{e^\frac{h}{y}}}^y=\frac{y}{\infty}^y=0$$

Answer 1

Apply L'Hospital's Rule to the $y$-th root of our function. So we want the limit of $\dfrac{h}{e^{h/y}}$. Now one step does it.

Answer 2

This is one of the fundamental limits associated with the exponential function and somehow there has been a belief that L'Hospital's rule is needed to prove it. However it should be understood that in most cases L'Hospital's rule is quite unnecessary. Rather we need to use some properties of exponential and logarithmic functions here.

Suppose that we know the series $$e^{h} = 1 + h + \frac{h^{2}}{2!} + \frac{h^{3}}{3!} + \cdots$$ (in fact this is taken as one of the definitions of $e^{h}$). Let $y > 0$ be given and let $n$ be an integer such that $n > y$. Then we know that $e^{h} > h^{n}/n!$ for $h > 0$ and hence $$0 < \frac{h^{y}}{e^{h}} < \frac{n!}{h^{n - y}}$$ Taking limits when $h \to \infty$ and noting that $n - y > 0$ we get $\lim_{h \to \infty}h^{y}/e^{h} = 0$.

Another approach is to take logs. Clearly $$f(h) = \log (h^{y}/e^{h}) = y\log h - h = yh\left(\frac{\log h}{h} - 1\right)$$ Now if we show that $(\log h)/h \to 0$ as $h \to \infty$ then we can see that because of $y > 0$ we have $f(h) \to -\infty$ and hence $h^{y}/e^{h} = e^{f(h)} \to 0$.

Next we handle the limit of $(\log h)/h$. Note that for $h> 1$ we can easily show that $\log h \leq h - 1$. This is equivalent to $e^{x} \geq 1 + x$ for all $x > 0$. This inequality of $\log$ or $\exp$ can be proven using any definition of $\log$ or $\exp$. Now let us assume that $h > 1$ and put $t = \sqrt{h}$ so that $t > 1$ and $h = t^{2}$. We can see that $$\log h = \log (t^{2}) = 2\log t \leq 2(t - 1) = 2(\sqrt{h} - 1)$$ Thus we have $$0 \leq \frac{\log h}{h} \leq \frac{2(\sqrt{h} - 1)}{h} = 2\left(\frac{1}{\sqrt{h}} - \frac{1}{h}\right)$$ Taking limits when $h \to \infty$ we get via squeeze theorem that $$\lim_{h \to \infty}\frac{\log h}{h} = 0$$