# Show ln x/x \< 1/e without derivatives?

**URL:** <https://boards.straightdope.com/t/show-ln-x-x-1-e-without-derivatives/356074>\
**Category:** Factual Questions\
**Created:** [May 9, 2006, 3:41pm UTC](https://boards.straightdope.com/t/show-ln-x-x-1-e-without-derivatives/356074 "2006-05-09T15:41:02Z")\
**Posts on this page:** 4\
**Page:** 2

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**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\
**Post date:** [May 12, 2006, 7:39pm UTC](https://boards.straightdope.com/t/show-ln-x-x-1-e-without-derivatives/356074/21 "2006-05-12T19:39:51Z")

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[QUOTE=Mathochist]  
you’re just reading in a statement that usually takes derivatives to prove, and thus doesn’t satisfy the requirements of the OP.  
[/QUOTE]

Strictly speaking, the OP said:

[QUOTE=panamajack]  
Without taking **the** derivative, can you show that (ln x)/x is always less than some value on [1, inf)  
[/QUOTE]

That is, he said we are not allowed to take the derivative of ln(x)/x, and not that we are not allowed to take any derivatives, or that we are not allowed to use calculus in any way.

(Of course the latter interpretation of the problem, i.e. solve without calculus, is more challenging and interesting)

On a related note, I once noticed that you could find the minimum of a quadratic function without using any derivatives.

Let f(x) = a_x^2 + b_x + c  
where a \> 0 (so f(x) has a minimum)

If you think about horizontal lines of the form  
g(x) = v  
for some value of v, the “highest” line that touches f(x)  
will correspond to the minimum of f(x).

i.e. we want to see what is the largest value of v s.t. f(x) = g(x) has a real solution.

```auto

f(x) = g(x)

==> a*x^2 + b*x + c = v

         -b +/- sqrt(b^2 - 4*a*(c-v))
==> x = ------------------------------
              2*a

```

We see that the above has a real solution only if  
the argument to the sqrt() is non-negative  
i.e. only if b^2 - 4_a_(c-v) \>= 0  
i.e. only if v \<= c - b^2/4\*a

So, the largest value of v that gives a real solution is c - b^2/4\*a  
and therefore this must be the minimum of f(x).

You can use the same method to find the minimum of any function whose zeros are easy to calculate.

An interesting aspect of this method is that it gives you the global minimum (or maximum) of a function, whereas the derivative method only gives you local minima and maxima, which you have to evaluate to see which is smallest (or largest)

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<div class="post-metadata">

**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\
**Post date:** [May 12, 2006, 7:43pm UTC](https://boards.straightdope.com/t/show-ln-x-x-1-e-without-derivatives/356074/22 "2006-05-12T19:43:36Z")

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[QUOTE=Polerius]  
On a related note, I once noticed that you could find the minimum of a quadratic function without using any derivatives.

[/QUOTE]

One final thing: the above method gives you not only the min, but also the arg min.  
When v is at its maximum value, the sqrt() becomes 0 and so the arg min is clearly seen to be x = -b/2\*a

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**Author:** ![Mathochist](https://avatars.discourse-cdn.com/v4/letter/m/c89c15/32.png) [@Mathochist](https://boards.straightdope.com/u/Mathochist)\
**Post date:** [May 12, 2006, 9:05pm UTC](https://boards.straightdope.com/t/show-ln-x-x-1-e-without-derivatives/356074/23 "2006-05-12T21:05:55Z")

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[QUOTE=Polerius]  
That is, he said we are not allowed to take the derivative of ln(x)/x, and not that we are not allowed to take any derivatives  
[/QUOTE]

It’s ludicrous to split that hair. If that was the intention, we could take the exponential of ln(x)/x, find _its_ derivative, and argue from there.

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<div class="post-metadata">

**Author:** ![Polerius](https://avatars.discourse-cdn.com/v4/letter/p/d78d45/32.png) [@Polerius](https://boards.straightdope.com/u/Polerius)\
**Post date:** [May 12, 2006, 9:27pm UTC](https://boards.straightdope.com/t/show-ln-x-x-1-e-without-derivatives/356074/24 "2006-05-12T21:27:54Z")

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[QUOTE=Mathochist]  
It’s ludicrous to split that hair.  
[/QUOTE]

I was partly joking 🙂

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