Show that $\lim_{x \to \infty} \frac{x^{10}2^x}{e^x}$ exists and is finite I want to show that $\displaystyle \lim_{x \to \infty} \frac{x^{10}2^x}{e^x}$ exists and is finite.
I have tried applying L'Hôpital's rule repeatedly, with no success. I cannot see a pattern or a way to simplify nicely.
I can obviously see that this integral evaluates to $0$, but have found no way to show it. Any hints?
 A: Guide:
From a single step of L'Hôpital's rule, notice that 
$$\lim_{x \to \infty} \frac{x^{10}}{\left( \frac{e}{2}\right)^x}= \lim_{x \to \infty} \frac{10x^{9}}{\ln\left( \frac{e}2\right)\left( \frac{e}{2}\right)^x}$$
Notice that the power of of the numerator reduces by $1$ and but the denominator form doesn't change much. What happens if you do it twice? what happens if you do it $10$ times?
A: Another option: for $k,\,n>0$, $\int_0^\infty x^n e^{-kx}=\frac{n!}{k^{n+1}}$ is finite, so $\lim_{x\to\infty}x^n e^{-kx}=0$.
A: $\dfrac{x^{10}2^x}{e^x} 
  = \dfrac{x^{10}e^{x \log 2}}{e^x}
  = \dfrac{x^{10}}{e^{(1-\log 2)x}}
  = \dfrac{x^{10}}{e^{Kx}}$
Where  $K = 1-\log 2 > 0$.
It shouldn't be too much trouble to show that the limit is $0$.
A: Consider $\displaystyle \lim e^{\log{\frac{x^{10}2^{x}}{e^{x}}}}$ , which equivalent to $e^{\lim (10\log{x} + x\log 2 - x)}$, so $\lim -x(\frac{10 \log{x}}{x} + (1-\log{2})) \to -\infty$, so you have $\lim e^{\dots} \to 0$.
A: Try this:
$\lim_{x \to \infty}(\frac{x^{10}2^{x}}{e^{x}}) = \lim_{x \to \infty}(\frac{x^{10}}{(\frac{e^{x}}{2^{x}})})$
Then use L'hopital here
A: The tenth root of the function is
$$xa^{-x}$$ where $a:=\sqrt[10]{\dfrac e2}>1.$
If you increase $x$ by $1$, this amounts to multiplying that expression by
$$c(x):=\frac{x+1}{ax}.$$
For $x>32.091\cdots$, this factor becomes smaller than $1$ so that the function can be bounded by a decreasing geometric sequence.
