Question

Without using any kind of derivatives, find \[\lim_{x\to\infty}x-\log_{10}(x).\]

Answer 1

What if you do \[
\left(\frac 1{10}\right)^{x−\log_{10}(x)}
\]

Answer 2

Hmmm...

Answer 3

\[
10^{\log_{10}(x)-x}=\frac{x}{10^x}
\]

Answer 4

can you use logic?

Answer 5

This question arose in a calc 1 class at a point where the students have not yet learned derivatives. We could probably use some basic logic, but not much more.

Answer 6

unless i am missing something, we know log grows slower than any positive power of \(x\)
much slower

Answer 7

put \(x=10^{12}\) get \(10^{12}-12\)

Answer 8

Hmm, I wonder if there is a clever way to factor out \(x\) from \(10^x\)

Answer 9

That's all fine and dandy, but how do we know it grows slow enough? If all I know about logs are the basic properties such as \(\log(ab)=\log(a)+\log(b)\), and that it goes to infinity as x goes to infinity, I don't see how to do it.

Answer 10

Hmmm, what about changing the base of the logarithm.

Answer 11

I suppose we could make an argument towards the fact that \(\log(x^2)=\log(x)+\log(x)\), and try and get some kind of inequality about \(x^2\) and \(\log(x)+\log(x)\).

You're welcome to try that. It ultimately shouldn't matter much.

Answer 12

Hmm \[
x>N\implies x-\log_{10}(x)>M
\]

Answer 13

We know it goes to infinity, so this is ultimately what we need to show.

Answer 14

It is unfair to say that a student would suspect it went to infinity, and then tried to prove it?

Answer 15

The closest I've come (in my opinion), is by assuming the fact that \[x\ge2\log_{10}(x)\]for \(x\) sufficiently large. Unfortunately, proving this inequality requires derivatives, unless I'm missing something clever.

Answer 16

Wouldn't it be for any arbitrary constant, not just 2, to show sufficiently large?

Answer 17

If we're given that \(a\ge2\log_{10}(a)\) for some \(a\), how do we know it's true for every \(b>a\)?

Answer 18

Okay how about this. \[\begin{split}
x-\log_{10}(x)&>M\\
10^{x-\log_{10}(x)}&>10^M\\
\frac{10^x}{x}&>10^M\\
\frac{2^x}{x}\times 5^x&>10^M\\
\end{split}
\]We know that \(2^x>x\) so \(2^x/x>1\)\[
\frac{2^x}{x}\times 5^x> 5^x>10^M
\]

Answer 19

Also \[
10^M=5^{M\log_5{10}}
\]

Answer 20

\[
x>N\implies 5^x>5^{M\log_5(10)}
\]

Answer 21

My only question about that, is can you show that \(2^x>x\) for \(x\) sufficiently large?

Answer 22

> is a lot easier to deal with than >>

Answer 23

Let me think about it...

Answer 24

Okay, so consider \(g(x)=2^x-x\).\[
\forall x_1,x_2>100\\
x_2>x_1\\
2^{x_2}>2^{x_1}\\
2^{x_2}-x_2>2^{x_1}-x_1
\]This means that \(g(x)\) is always increasing.

Answer 25

It shows:\[
x_2>x_1\implies g(x_2)>g(x_1)
\]Meaning \(g(x)\) is an increasing function for arbitrarily large \(x>100\)

Answer 26

Not only is it always increasing, but it is positive beyond \(100\).

Answer 27

So if \(2^x\) is greater than \(x\) at 100 and beyond 100 the gap never narrows, then that means for arbitrarily large \(x\) we know \(2^x>x\)

Answer 28

Is it still a bit iffy?

Answer 29

It's looking better. Give me a moment to verify everything.

Answer 30

A lot of the inequality stuff only holds for arbitrarily large \(x\). Not for small or negative \(x\) obviously.

Answer 31

Since we're looking at \(x\to\infty\), we can disregard small or negative \(x\).

Can you explain how you concluded that the gap never narrows?

Answer 32

Actually, I don't think that matters. We just need \(2^x>x\) for sufficiently large \(x\), and I think this does it.

Answer 33

The gap between \(2^x\) and \(x\) is \(2^x-x\).

Answer 34

This gap has to be \(0\) for one to overtake the other.

Answer 35

Otherwise whoever is greater will retain supremacy.

Answer 36

Excellent work. I think this proves the limit without resorting to derivatives.