$\lim a^x x$ when $x$ approaches $\infty$ If $|a|<1$, show that
$$\lim_{x\to \infty} a^x x =0$$
L'Hospital's Rule doesn't work, in this case. How can I analyze the given function to conclude this ?
 A: If $a=0$, the claim is trivial. Else, define $b=\frac{1}{a}$. Because absolute convergence implies convergence, you might assume $a>0$ so that $b>1$. Now
$$
a^xx=\frac{x}{b^x}
$$
So L'Hospital's Rule actually applies
$$
\lim_{x\to\infty}a^xx=\lim_{x\to\infty}\frac{x}{b^x}=\lim_{x\to\infty}\frac{1}{b^x\log b}=0.
$$
A: Rewrite
$$\lim_{x\to \infty} a^x x =\lim_{x\to \infty} \frac{x}{a^{-x}}$$ 
Now use L'Hospital
$$ = \lim_{x\to \infty} \frac{1}{-a^{-x}\ln a} = \lim_{x\to \infty} \frac{-a^x}{\ln a} = 0$$ 
A: $$xa^x = \frac{x}{\left(\frac{1}{a}\right)^x}$$
now apply l'Hospital.
A: Note that $\ln a ^x x = x\ln a +\ln x$. It follows that
$$
\frac{\ln a^xx}{x}\rightarrow \ln a
$$
as $x\rightarrow \infty$. Now consider what happens to $\ln a^xx$ and apply the exponential function.
A: Hint
I suggest that you consider the different cases :$a>1$,$a=1$,$a<1$. For the first two cases, the answer seems to be obvious :$\infty$. But the problem is totally different when $a<1$.
I am sure that you can take from here.
A: I'm just wondering if L'Hospital's Rule works for the anti-derivative. For example, one would like to evaluate
$$\lim_{x\to \infty} a^x x = \lim_{x\to \infty} \frac{a^x}{\frac{1}{x}}=\lim_{x\to \infty} \frac{a^x}{\frac{1}{x}}=\lim_{x\to \infty} \frac{\int a^xdx}{\int\frac{1}{x}dx}=\lim_{x\to \infty} \frac{\frac{a^x}{\ln a}}{\ln \rvert x\lvert}=0$$
Because the numerator tends to $0$ and the denominator tends to $\infty$, the limit tends to $0$.
Is this valid?
