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I need to prove this limit using Heine definition, given $a \in \mathbb R$ and $b > 1$. $$ \lim_{x\to \infty} \frac {x^a} {b^x} = 0 $$ How can this be done?

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    Hint: An exponential grows faster than a finite polynomial. – 2017-01-02
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    Thanks, I do understand, in terms of growth, that this limit is 0. I can't find way to manipulate the expression. – 2017-01-02
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    What is Heine definition? – 2017-01-02
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    [Heine Definition](https://en.wikipedia.org/wiki/Limit_of_a_function#In_terms_of_sequences) – 2017-01-02
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    What's your definition of $b^x$ or $a^x$? – 2017-01-03

2 Answers 2

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We want to prove that for any sequence $(x_n)$ such that $x_n\to +\infty$, we have that $\frac{x_n^a}{b^{x_n}}\to 0$. That is,

$$(\forall\varepsilon>0)(\exists n_0\in\Bbb N)\ n\geq n_0\implies\frac{x_n^a}{b^{x_n}}<\varepsilon$$

We have that $\frac{x_n^a}{b^{x_n}}<\varepsilon\iff a\ln x_n - x_n\ln b <\ln\varepsilon$. That is, it is enough to prove that $$a\ln x_n-x_n\ln b\to -\infty$$

which is true because $\lim_n\frac{\ln x_n}{x_n} = 0$:

$$\lim_n(a\ln x_n-x_n\ln b) = \lim_n x_n(a\frac{\ln x_n}{x_n}-\ln b) = [\ln b > 0] = -\infty.$$

What remains to be proved is that $\frac{\ln x_n}{x_n}\to 0$, that is

$$(\forall\varepsilon>0)(\exists n_0\in\Bbb N)\ n\geq n_0\implies\frac{\ln x_n}{x_n}<\varepsilon.$$

But, $$\frac{\ln x_n}{x_n}<\varepsilon\iff \frac{e^{\varepsilon x_n}}{\varepsilon x_n}>\frac 1{\varepsilon}$$

and since $\varepsilon x_n\to +\infty$, it only remains to prove $\lim_{x\to +\infty}\frac{e^x}{x} = +\infty.$

How to prove last statement depends on how you define $e^x$, but you can use $e^x\geq 1 + x + \frac {x^2}2$ when $x>0$, for example.

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Hint

Taking logarithm, we get

$$\ln(f(x))=$$ $$g(x)=a\ln(x)-x\ln(b)$$

$$=x(a\frac{\ln(x)}{x}-\ln(b))$$

using the known limit $\lim_{x\to +\infty}\frac{\ln(x)}{x}=0$ and $\ln(b)>0$,

we find $\lim_{x\to+\infty}g(x)=-\infty$

and $\lim_{x\to +\infty}f(x)=0$.