Proving the Limit of log(x+1)/x

[Akorys]

Homework Statement

Prove the limit \lim_{x\rightarrow 0} \frac{\log(x+1)}{x} =1

Homework Equations

Use the relation 1 - \frac{1}{x} \leq \ \log x \leq x-1\ \text{if}\ x>0

The Attempt at a Solution

We need to show that |\frac{\log(x+1)}{x} - 1 | \lt \epsilon\ \text{whenever}\ 0 \lt |x| \lt \delta. Using the relation above, we find \log x \lt x \frac{\log(x+1)}{x} \lt \frac{x+1}{x}
We can prove the right-hand limit by considering only those x \gt 0, since in that case \frac{\log(|x|+1)}{|x|} = |\frac{\log(x+1)}{x}|. |\frac{\log(x+1)}{x}-1| \leq |\frac{\log(x+1)}{x}| + 1 \lt |\frac{x+1}{x}| +1 = \epsilon
We need a \delta\ \text{such that} \ 0 \lt|x|\lt\delta. If \delta = \frac{1}{\epsilon -1} =|\frac{x}{x+1}| \lt |x|, then this proves the right-hand limit. However, I do not know how to account for -1\lt x \lt 0, since \log(x+1) is defined for such x and thus has a left-hand limit.

Also, please let me know if my current proof is flawed since I do not completely understand how to do epsilon-delta proofs. Thank you!

 

[GFauxPas]

I'm not sure what you're doing here:

Using the relation above, we find \log x \lt x \frac{\log(x+1)}{x} \lt \frac{x+1}{x}

but you don't need to do an epsilon delta proof for this one, there's a much much simpler way.

What do you know about the limit of a function that's squeezed between two other functions?

 

[Akorys]

I can explain. Since \log x \leq x-1 \lt x\ \text{for}\ x \gt 0 \text{, then}\ \log(x+1) \lt x+1 \text{, }\ x \gt -1. But yes, the squeezing principle, that worked. It indeed works out to 1. Thank you for the advice!

 

[Mark44]

I'm not sure what you're doing here:

but you don't need to do an epsilon delta proof for this one, there's a much much simpler way.

The "prove" in the problem statement might mean that an epsilon-delta proof is required, although the inequalities in the relevant equations section suggests that an epsilon-delta proof isn't what is being asked for.

What do you know about the limit of a function that's squeezed between two other functions?

 

[Akorys]

Thank you all for the advice. Although I most likely did not need an epsilon delta proof, I still wrote one up that I'm fairly confident in. I can post it for review, but I used the squeezing principle to prove the limit, as well as the fact that (log 1)' = 1, and so I don't want to take up people's time with proofs of problems I already know some solutions to.

 

[Mark44]

... as well as the fact that (log 1)' = 1

I don't know what you mean by this. The derivative of any constant is zero.

 

[Akorys]

Oh yes I meant if you take (log x)' then that's (1/x). Then when x = 1, the derivative is 1.

Using this fact, you can use the definition of the derivative as [log(1+x) - log(1)]/x as x tends to 0. Log(1) = 0, so you're left with the expression I was trying to work with:
[log(x+1)]/x as x tends to 0.

From that we can say that the limit is equal to the derivative of log at 1, which is 1. Sorry for the confusion, and I can explain anything else I wrote that may seem/be flawed.

 

[GFauxPas]

Thank you all for the advice. Although I most likely did not need an epsilon delta proof, I still wrote one up that I'm fairly confident in. I can post it for review, but I used the squeezing principle to prove the limit, as well as the fact that (log 1)' = 1, and so I don't want to take up people's time with proofs of problems I already know some solutions to.

I'd like to see it.

 

[Akorys]

Then I'd appreciate any criticism you may have of it.

We are given the relation 1-\frac{1}{x} \leq \log x \leq x-1, for x \gt 0. Moreover, the equality only holds for x=1 (this was proved in another question). Then we can substitute x by (x+1) and attain; 1-\frac{1}{x+1} \leq \log (x+1) \leq x \text{, }\ x \gt -1
Now we would have a strict equality for x = 0. Consider only those x>0. Then we have; \frac{\log(x+1)}{x} \lt 1
Since the expression on the left is positive for all values of x>0, we have |\frac{\log(x+1)}{x}| \lt 1

Now consider only those x such that -1<x<0. Then we have; \frac{\log(x+1)}{x} \lt \frac{1-\frac{1}{x+1}}{x}
This reduces to: \frac{\log(x+1)}{x} \lt \frac{1}{x+1}
Since both the expressions are positive for -1<x<0, we can state that |\frac{\log(x+1)}{x}| \lt |\frac{1}{x+1}|

Having found these results, we now turn to the proof of the limit. We must show that |\frac{\log(x+1)}{x} -1| \lt \epsilon \ \text{whenever }\ 0 \lt |x| \lt \delta
By the triangle inequality, we find that |\frac{\log(x+1)}{x} -1| \leq |\frac{\log(x+1)}{x}| + 1

We can complete the proof by taking x>0 first, and then -1<x<0, substituting the expression by the results we obtained from the given relation. For x>0; |\frac{\log(x+1)}{x}| + 1 \lt 1+1 = 2 \lt \frac{1}{|x|} = \epsilon
We let \epsilon = \frac{1}{|x|} since \frac{1}{|x|} \gt 2 \text{ if }\ |x| \lt \frac{1}{2}, hence this ensures \epsilon is always greater than the absolute value expression as x tends to 0. We then take delta as follows: 0 \lt |x| \lt \delta = |x| +1 =\frac{1}{\epsilon} -1 = \frac{1-\epsilon}{\epsilon}.
This completes the proof for x>0. The proof for -1<x<0 is analogous. From the results of the triangle inequality, using the results from the original relation, we obtain |\frac{\log(x+1)}{x}| + 1 \lt |\frac{1}{x+1}| + 1 = \epsilon
Now we obtain delta: 0 \lt |x| \lt \delta = |x+1| = \frac{1}{\epsilon -1}
This completes the proof.

And again, I should restate that I am not very comfortable with these types of proofs, so if my method is completely flawed in some way then please point it out. Thank you in advance for all comments on this!

Just as an aside, I can see that my method is most likely flawed, since it seems to be able to prove the limit for any solution, at least as far as I can see. I am simply unsure of what we can and cannot do/assume when actually creating the proof.

Edit* As far as I understand, the flaw in my "proof" is that I cannot choose the epsilon to be what I want it to be, but instead it must be based on any epsilon greater than zero. I apologize for wasting people's time with some of my work, and would like to ask if this is indeed the issue?

 

[Mark44]

There's no problem with assuming that $\epsilon$ be positive.

 

[GFauxPas]

Unless I'm wrong, it looks like your proof only works for $\epsilon > 2$.
Don't apologize for wasting our time - there's no requirement for us to read your post, after all!