When $x$ tends to infinity then find limit Suppose $$f(x)=(\ln(x))^2$$ 
Then as $x$ tends to $\infty$, find the limit of 
$$f(x+1)-f(x)$$
(if it exists).
$$\begin{align}
f(x+1)-f(x)&=(\ln(x+1))^2-(\ln(x))^2 \\
&=(\ln(x(1+1/x)))^2-(\ln(x))^2 \\
&=2\ln(x)\ln(1+1/x)+((\ln(1+1/x))^2
\end{align}$$
Then what can I do?
 A: I think this limit is zero. Here is the justification:
$$L = f(x+1)-f(x) = \ln(x+1)^2-\ln(x)^2 = (\ln(x+1)-\ln(x))(\ln(x+1)+\ln(x))$$
This is a $0 \times \infty$  form. Let's make it $\frac{0}{0}$ so we can use L'Hospital's rule.
$$L = \frac{\ln(x+1)-\ln(x)}{\frac{1}{\ln(x+1)+\ln(x)}}$$
Now, differentiate both numerator and denominator and the limit will stay the same.
$$L = \frac{\frac{1}{x+1}-\frac{1}{x}}{\frac{-1}{(\ln(x+1)+\ln(x))^2}\left(\frac{1}{x+1}+\frac{1}{x}\right)}$$
Simplifying we get:
$$L = \frac{(\ln(x+1)+\ln(x))^2}{2x+1}$$
Now, apply L'Hospitals rule and differentiate numerator and denominator again.
$$L = (\ln(x+1)+\ln(x))\left(\frac{2x+1}{x(x+1)}\right)$$
As $x \to \infty$ this becomes equivalent to:
$$L = \frac{2(\ln(x+1)+\ln(x))}{x}$$
It's clear that the denominator increases much faster than the nuumerator and the limit must be $0$. 
If not, just apply L'Hospital's rule again:
$$L = \lim_{x \to \infty}2\left(\frac{1}{x+1}+\frac{1}{x}\right)$$
This is obviously $0$.
Note that this makes sense since for large $x$, your $f(x)$ should asyptotically become quite stable with very large jumps needed to increase it in any meaningful way. So, a jump of $1$ won't do much to move it.
A: $$\Delta =f(x+1)-f(x) = \log(x+1)^2-\log(x)^2 = (\log(x+1)-\log(x))\,(\log(x+1)+\log(x))$$
$$\log(x+1)-\log(x)=\log \left(1+\frac{1}{x}\right)$$
$$\log(x+1)+\log(x)=\log \left(1+\frac{1}{x}\right)+2\log(x)$$
Use Taylor
$$\log \left(1+\frac{1}{x}\right)=\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}+O\left(\frac{1}{x^4}\right)$$
$$\Delta =\left(\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}+O\left(\frac{1}{x^4}\right) \right)\left(\frac{1}{x}-\frac{1}{2 x^2}+\frac{1}{3
   x^3}+O\left(\frac{1}{x^4}\right)+2\log(x) \right)$$
$$\Delta \sim\frac{2\log (x)}{x}$$
A: Another user has already supplied a of rigorous proofs the limit is 0, but it's worth mentioning that a good first step is to simplify the prblem as much as possible first. You've made good progress, but there's one more step I would take.
In particular, as you look at the problem in the form you've got it in already, that is, $f(x+1) - f(x) = 2\ln(x)\ln(1+1/x)+(\ln(1+1/x))^2$, you can see that there are a few places where you have the term $\frac{1}{x}$. As you take $x$ to infinity, the righthand term $(\ln(1+1/x))^2$ obviously goes to $0$ (since limits can distribute across sums and inside squares so that $\lim_{x\to \infty}(\ln(1 + 1/x))^2 = \ln(1)^2 = 0$), so you can effectively ignore this term, and you are left with the limit of the lefthand term, $\lim_{x \to \infty} 2\ln(x)\ln(1 + 1/x)$. 
This leaves you with the (relatively simpler) job of determining $\lim_{x \to \infty} 2\ln(x)\ln(1 + 1/x)$. I'm not sure how rigorous a proof you're looking for, but this gives you a much easier problem to work with (assuming you're being allowed to calculate limits like we did in the first paragraph). I won't waste space making this final calculation for you, since I'm sure you know how to take limits of this sort, and can finish the calculation on your own.
