I was trying to calculate $\lim_{x \to 0^{+}} x^x$ without L'Hôpital's rule but could not make progress.
My best shot was to show that $\lim_{x \to 0^{+}} x\ln x = 0$ as that would imply the first limit. Can anyone help me?
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1Why without l'Hospital rule? – Berci Mar 06 '16 at 01:17
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1I´m studying for test where I will not be allowed to use it. – Gabriel Mar 06 '16 at 01:19
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1Ah, ok. Next step might be to use $x=e^t$ then we look for $e^t\cdot t$ while $t\to -\infty$. – Berci Mar 06 '16 at 01:20
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Hint: What else do you know about $\ln x$? How does its growth rate compare to $\frac1x$ or $x$?
Another hint: you can transform this to a $\lim_{y \rightarrow \infty}$ problem i.e. by setting $y=1/x$. I find this much easier to work with. Thinking about things close to $0^+$ is hard.
You will want to use some combination of finding upper / lower bounds, and technically you will apply the squeeze theorem.
djechlin
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I´m trying to show that ln(x) grows asymptotically slower than x. That is, $\lim_{y \to \infty} \frac{\ln y}{y} = 0$. – Gabriel Mar 06 '16 at 01:30
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1@GabrielRibeiro that's well known. L'H is one way. Consider the sequence $e, e^2, e^3, \ldots$. – djechlin Mar 06 '16 at 01:33
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To do this from scratch, here's one approach:
First note that $x\ln x<0$ as soon as $x<1$.
Next, observe that $ne^{-n}\to 0$ as $n\to \infty ,\ $ because $ne^{-n}<\frac{n}{1+n+\frac{n^{2}}{2}}$.
And since $(x\ln x)'=\ln x+1,\ h(x)=x\ln x$ is decreasing on $(0,e^{-n})\ $ for all integers $n\geq 1$, which means we must have $x\ln x>-ne^{-n}$ for all $x\in (0,e^{-n})$.
Putting this all together then, we can write
$-ne^{-n}<x\ln x<0$ whenever $x\in (0,e^{-n})$ and now to finish, we need only appeal to the squeeze theorem to obtain the result.
Matematleta
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In THIS ANSWER I showed using on the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\frac{x-1}{x}\log(x)\le x-1 \tag 1$$
for $x>0$.
Using $\log(x^{\alpha})=\alpha \log(x)$ for all $\alpha$, we have for any $\alpha >0$ and $x>0$
$$\frac{x-x^{1-\alpha}}{\alpha}\le x\log(x)\le \frac{x^{1+\alpha}-x}{\alpha}$$
whereupon choosing $\alpha <1$ and applying the squeeze theorem reveals
$$\lim_{x\to 0^+}\left(x\log(x)\right)=0$$
Finally, we have
$$\begin{align} \lim_{x\to 0^+}x^x&=\lim_{x\to 0^+}e^{x\log(x)}\\\\ &=e^{\lim_{x\to 0^+}\left(x\log(x)\right)}\\\\ &=1 \end{align}$$
TOOLS USED:
(i) The limit definition of the exponential function;
(ii) Bernoulli's Inequality;
(iii) $\log(x^{\alpha})=\alpha \log(x)$;
(iv) The Squeeze Theorem;
(v) Continuity of the exponential function;
(vi) $\lim_{x\to a}f(g(x))=f\left(\lim_{x\to a}g(x)\right)$ when $f$ is continuous.
Mark Viola
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Consider $A=x^x$. Taking logarithms $\log(A)=x\log(x)$ which goes to $0$ if $x\to 0$. So $\log(A)\to 0 \implies A \to 1$.
Claude Leibovici
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