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 $\alpha \rightarrow \infty$


  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.

  • 1
    $\begingroup$ 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


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!

  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ @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

Hint. Use differentiation to show that $$f(x)=\frac{\ln x}{x^a}$$ has a maximum value of $\dfrac1{ae}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.