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Question: Assume $(\ln x)' = \frac1x$, $(x^\alpha)' = \alpha x^{\alpha-1}$ for all $x>0$, $\ln(e^x)=x$ and that $\alpha>0$.

  1. Prove that $\ln x \leq x^\alpha$ for large x.
  2. Prove that there exists a constant $c_\alpha$ such that:

    (a) $\ln x \leq c_\alpha x^\alpha$ for all $x \in [1,\infty)$

    (b) $c_\alpha \rightarrow \infty$ as $\alpha \rightarrow 0^+$

    (c) $c_\alpha \rightarrow 0$ as $\alpha \rightarrow \infty$

Attempt:

  1. Consider $f(x) = \frac{\ln x}{x^\alpha}$. Then $$\lim\limits_{x\to\infty}f(x) \stackrel{\text{L'H}}{=} \lim\limits_{x\to\infty}\frac{1}{\alpha x^\alpha}=0.$$ This implies that $x^\alpha$ grows at a faster rate than $\ln x$. Therefore, $\ln x \leq x^\alpha$ for large x.

  2. I'm thinking that if a function is greater than another, then the derivative of the larger function should be greater than (or equal to) the derivative of the smaller function as x gets large. Therefore, taking the derivatives and solving for x we get: $$\frac1x = \alpha x^{\alpha-1} \Rightarrow x = \left( \frac1\alpha \right)^{\frac1\alpha}.$$ So I thought I should let $c_\alpha = \left( \frac1\alpha \right)^{\frac1\alpha} - 1$ so that it satisfies conditions b and c, but it does not satisfy condition a. I need a hint.

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    $\begingroup$ Your statement (2) is not correct. For example if $f(x)=x+x^{-1}$ and $g(x)=x$ then $f(x)>g(x)$ for all $x>0$, but $f'(x)<g'(x)$. Draw graphs of $f$ and $g$ on the same axes and you will understand what is happening here. $\endgroup$ – David Nov 14 '16 at 4:14
  • $\begingroup$ @David: Thanks. I am looking at a graph of both functions. $f'(x) \rightarrow g'(x)$ as $x \rightarrow \infty$. This was the kind of idea I thought would be useful. $\endgroup$ – SOULed_Outt Nov 14 '16 at 4:20
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PRIMER:

In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that

$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1 }\tag 1$$

for $x>0$.

Aside, I showed in THIS ANSWER, using elementary (pre-calculus) tools, that for all positive integers, $\log(n) \le \sqrt{n}$. In fact, $\log(x)\le \sqrt{x}$ for all $x>0$.


Note that $\log(x)\le x-1<x$ for $x>0$. Then, $\alpha \log(x)=\log(x^\alpha)\le x^\alpha$. If $\alpha >0$, then we have

$$\log(x)\le \frac{x^\alpha}{\alpha}$$

And we are done!

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  • $\begingroup$ Is it safe to say that for any real-valued function, say $f$, the following inequality holds: $$\log(f) \leq f-1$$ $\endgroup$ – SOULed_Outt Dec 3 '16 at 2:25
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    $\begingroup$ Yes, if $f>0$, then $\log(f)\le f-1<f$. $\endgroup$ – Mark Viola Dec 3 '16 at 2:27
  • $\begingroup$ How should I go about proving the inequality holds for any positive real-valued function? $\endgroup$ – SOULed_Outt Dec 3 '16 at 2:35
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    $\begingroup$ @SOULed_Outt I've added a primer with a link to an answer I posted HERE using elementary tools only. -Mark $\endgroup$ – Mark Viola Dec 3 '16 at 2:39
  • $\begingroup$ @SOULed_Outt You're welcome. I also showed in THIS ANSWER, using elementary (pre-calculus) tools, that for all positive integers, $\log(n) \le \sqrt{n}$. And in fact, $\log(x)\le \sqrt{x}$ for all $x>0$. $\endgroup$ – Mark Viola Dec 3 '16 at 3:03
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Hint. Use differentiation to show that $$f(x)=\frac{\ln x}{x^a}$$ has a maximum value of $\dfrac1{ae}$.

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