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How can I compute $\lim_{x \to \infty} \frac{e^x}{x^a}$ for some $a \in \mathbb R$ with $x^a := e^{a \log(x)}$?

I want to use only the basic properties of limits, i.e. the linearity, multiplicativity, monotonicity and the Sandwich property (no L'Hospital).

Can you give me a hint?

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  • $\begingroup$ How about the continuity of $\ln$? $\endgroup$ – Michael Burr Jan 5 '18 at 13:27
  • $\begingroup$ Perhaps someone can provide a more rigorous proof, but $e^x$ grows alot faster than the denominator.$e^x = 1 + x + \frac{x^2}{2} + ..$ $\endgroup$ – Rick Jan 5 '18 at 13:29
  • $\begingroup$ @MichaelBurr That's ok of course. Just no derivatives, L'Hospital etc. $\endgroup$ – philmcole Jan 5 '18 at 13:29
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    $\begingroup$ Did using$ e^{x}x^{-a}$ then using the formula get you anywhere? $\endgroup$ – smokeypeat Jan 5 '18 at 13:42
  • $\begingroup$ @smokeypeat Thanks, that helps. By continuity of $\exp(x)$ I then get $\exp \big( \lim_{x \to \infty} (x- a \log(x)) \big)$. Though how can I show simply that $\lim_{x \to \infty} x- a \log(x)$ diverges? $\endgroup$ – philmcole Jan 5 '18 at 15:19
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another method to proof it is with the definition of $e^x$ as $e^x=\sum_{n=0}^{\infty}\frac{z^n}{n!}$

Take $n \in \mathbb{N}$ such that $a<n$ than it follows that

$$ \frac{e^x}{x^{a}} > \frac{e^x}{x^n} = \frac{1}{x^n} \sum_{i=0}^{\infty}\frac{x^{i}}{i !} > \frac{1}{x^n}\frac{1}{(n+1)!}x^{n+1}=\frac{x}{(n+1)!} \longrightarrow \infty $$ for $x \longrightarrow\infty$.

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you may find the following line of thought helpful.

choose $x_0$ so that $e^{x_0}\gt 2^a$ and write $f(x)$ for $e^x$ and $g(x)=x^a$, so that $$ h(x) = \frac{e^x}{x^a} = \frac{f(x)}{g(x)} $$

then $$ h(2x_0)= \frac{f(x_0)^2}{2^a g(x_0)} = Kh(x_0) $$ where $K=\frac{f(x_0)}{2^a} \gt 1$ and then we have $$ h(2^nx_0) = K^nh(x_0) $$

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Here is an entirely elementary proof. First we have that if $a>1$ then

$$n+1\leq a^n$$ for large enough $n$. This follows from the binomial theorem,

$$(a-1)^2\frac{n(n-1)}{2} < (1+(a-1))^n$$

now chose $n$ large enough so that $$n+1 < (a-1)^2\frac{n(n-1)}{2}$$ which is clearly possible.

Next, $$x < a^x$$ for large enough $x$, for if $n\leq x < n+1$ then

$$x <n+1 < a^n <a^{x} $$

Now it follows that

$$x^k < a^x$$ for all large $x$. Just let $a=b^k$, then

$$x < b^x$$ and taking the power $k$,

$$x^k < a^x.$$

It follows that given $k$,

$$x^{k+1} < a^x$$ and so

$$x<\frac{a^x}{x^k}$$ and the limit is infinity.

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