Proving $\lim_{x \to +\infty}\frac{\ln(x)}{2\cdot \ln(x+1)} = \frac{1}{2}$ with $\epsilon- \delta$

Question

Today, I spent some hours trying to solve this problem but I can't reach the conclusion:

Show with $\epsilon-\delta $ definition the following limit: $$\lim_{x \to +\infty}\frac{\ln(x)}{2\cdot \ln(x+1)} = \frac{1}{2}$$

So, first of all, I set up the inequality: $$\left|\frac{\ln(x)}{2\cdot \ln(x+1)}-\frac{1}{2}\right|<\epsilon$$

Then: $$\frac{1}{2}-\epsilon<\frac{\ln(x)}{2\cdot \ln(x+1)}<\frac{1}{2}+\epsilon \implies (1-2\epsilon)\cdot \ln(x+1) < \ln(x) < (1+2\epsilon)\cdot \ln(x+1)$$

And elevating with base $e$ to cancel the $\ln(x)$:

$$(x+1)^{1-2\epsilon}<x<(x+1)^{1+2\epsilon}$$

I want to apply the limit $\lim_{x \to +\infty}\frac{(1+f(x))^\alpha-1}{\alpha\cdot f(x)}=1$ with $f(x) \to_{x \to 0} 0$. So, I have:

$$x^{1-2\epsilon}\cdot\left(1+\frac{1}{x}\right)^{1-2\epsilon}<x<x^{1+2\epsilon}\cdot\left(1+\frac{1}{x}\right)^{1+2\epsilon}$$

Then:

$$ \left\{\begin{matrix} x>x^{1-2\epsilon}\cdot\left(1+\frac{1}{x}\right)^{1-2\epsilon} \\ x<x^{1+2\epsilon}\cdot\left(1+\frac{1}{x} \right )^{1+2\epsilon} \end{matrix}\right. $$

So: $$ \left\{\begin{matrix} x^{2\epsilon} - 1>\left(1+\frac{1}{x}\right)^{1-2\epsilon} - 1 \\ \frac{1}{x^{2\epsilon}}-1<\left(1+\frac{1}{x} \right )^{1+2\epsilon}-1 \end{matrix}\right. $$ When $x$ approaches $+\infty$: $$ \sim \left\{\begin{matrix} x^{2\epsilon}-1>\frac{1-2\epsilon}{x} \\\frac{1}{x^{2\epsilon}}-1<\frac{1+2\epsilon}{x} \end{matrix}\right. $$

Notice that the second inequality is always true because $\frac{1}{x^{2\epsilon}}\to 0^+$ when $x \to +\infty$, so that left side is negative while right side $\frac{1-2\epsilon}{x}$ is positive.

Now, in some way I have to find $\delta = N$. I noticed that: $$x^{2\epsilon}-1\sim x^{2\epsilon} \;\;\; x \to +\infty$$

and then for the first inequality: $$x^{2\epsilon}>\frac{1-2\epsilon}{x} \implies x> \delta(\epsilon)= N =(1-2\epsilon)^{1+2\epsilon}$$

Here, tere is a problem because when $\epsilon \to 0^+$, $\delta = N \to 1$ and not to $+\infty$. Is this anyway correct? Or, I can't apply asympothics reduction with $\epsilon - \delta$?

Edit: I all three solutions proposed, $\delta = N$ goes to infinity when $\epsilon \to 0^+$. In my answer, $N$ tends to $1$, so is my solution not corret? If it's yes, why?

Thanks.

Answer 1

To prove that $\log_{x+1}x\stackrel{x\to+\infty}\to1$, we need to find $N$ such that $|\log_{x+1}x-1|<\epsilon$ for all $\epsilon>0$ and $x>N$. This is equivalent to $(x+1)^{1-\epsilon}<x<(x+1)^{1+\epsilon}$ where the upper bound is obvious. The lower bound is also obvious when $\epsilon\ge1$ (take $N>1$), and when $0<\epsilon<1$, Bernoulli's inequality gives $(x+1)^{1-\epsilon}\le1+(1-\epsilon)x$ so taking $N>1/\epsilon$ completes the proof.

Answer 2

I'm sorry I had not the patience to read through all what you did but here is a way to prove the limit with elementary arguments: the intuition is indeed that $\ln(x)$ and $\ln(1+x)$ are big and close to each other. Instead of calculating the ratio, let us look at the difference:

$$\ln(1+x) - \ln(x) = \ln\left( \frac{1+x}{x}\right)= \ln\left(1+ \frac{1}{x}\right) < \frac{1}{x}\quad \text{for}\quad x > 0$$ (graph of $\ln(1+t)$ below its tangent at $t=1$ because the function is concave. Can also be seen with the analytic expansion). From here, one has $$ 1 - \frac{\ln(x)}{\ln(1+x)} < \frac{1}{x\, \ln(1+x)}\quad \Longrightarrow\quad 1 - \frac{1}{x\, \ln(1+x)} < \frac{\ln(x)}{\ln(1+x)} $$ (and r.h.s. is obviously smaller than $1$). Since $x\, \ln(1+x)\to \infty$ when $x\to\infty$, the inverse $\frac{1}{x\, \ln(1+x)} \to 0$

Answer 3

Fix $x>e^2-1$. Using $$ \ln(1+\frac1x)<\frac1x $$ we have $$ \left|\frac{\ln(x)}{2\cdot \ln(x+1)}-\frac{1}{2}\right|=\frac12\frac{\ln(1+\frac1x)}{\ln(x+1)}<\frac{1}{4x}. $$ Then for $\forall \varepsilon>0$, denote $N$ by $N=\max\{e^2-1,\frac1{4\varepsilon}\}$. Then when $x>N$, $$ \left|\frac{\ln(x)}{2\cdot \ln(x+1)}-\frac{1}{2}\right|<\varepsilon. $$