Limit at infinity $\lim_{x\to \infty} x^a a^x=$?

Question

$\displaystyle \lim_{x\to \infty} x^a a^x=?$; $0<a<1$

I try to use the property: $a^{\log_a x}= x$ and reescribe the expression

$\displaystyle \lim_{x\to \infty} x^a a^x = \lim_{x\to \infty} \frac{a^{a\log_a x}}{a^{-x}} $ but i can't find the limit yet.

Any suggestion?

Answer 1

Consider $x=ay$; then your expression becomes $$ x^aa^x=(ay)^aa^{ay}=a^a(ya^y)^a $$ so you just need to compute $$ \lim_{y\to\infty}ya^y $$ For $a\ge1$ the limit is $\infty$. For $0<a<1$ rewrite it as $$ \lim_{y\to\infty}\frac{y}{a^{-y}}=\lim_{y\to\infty}\frac{1}{-a^{-y}\log a}=0 $$ with a simple application of l'Hôpital.

Answer 2

HINT

For $a\ge 1$ it is trivial.

For $0<a<1$ let $b=\frac1a>1$ then

$$x^aa^x=\frac{x^a}{b^x}$$

then we can easily conclude for example by l’Hopital.

Answer 3

Take the logarithm: $\ln x^a a^x = a \ln x \, x \ln a = (a \ln a) (x \ln x)$

Here, $x \ln x \to \infty,$ but the factor $a \ln a$ can change the limit.

If $a>1$ then $a \ln a > 0$ so $(a \ln a) (x \ln x) \to \infty$ making $x^a a^x \to \infty.$

If $a=1$ then $a \ln a = 0$ so $(a \ln a) (x \ln x) \to 0$ making $x^a a^x \to 1.$

If $0<a<1$ then $a \ln a < 0$ so $(a \ln a) (x \ln x) \to -\infty$ making $x^a a^x \to 0.$