How to evaluate $\lim_{x\to\infty}\left(\frac{x+n}{\ln(x+n)}-\frac x {\ln x}\right)$? The question is somewhat related to the discussions I had about this other post. It seems for every integer $n&gt;1$ the limit
$$\lim_{x\to\infty}\left(\frac{x+n}{\ln(x+n)}-\frac x {\ln x}\right)$$
does exist. But I wasn't able to obtain a closed form for it (I admit that I didn't try enough). So, is there a closed form for it?
Hint. By the Mean Value Theorem,
$$\frac{x+n}{\ln(x+n)}-\frac{x}{\ln x}=((x+n)-x)\cdot\frac{d}{dx}\left(\frac{x}{\ln x}\right)_{x=t}=n\left(\frac{1}{\ln t}-\frac{1}{\ln^2 t}\right)$$
with $x&lt;t&lt;x+n$. Now $t$ goes to infinity as $x\to\infty$.
see my answer
$$\lim_{x\to\infty} \frac{x+n}{\ln(x+n)}-\frac{x}{\ln x}$$
$$=\lim_{x\to\infty} \frac{x\left(1+\frac nx\right)}{\ln x\left(1+\frac{n}{x}\right)}-\frac{x}{\ln x}$$
$$=\lim_{x\to\infty} \frac{x\left(1+\frac nx\right)}{\ln x+\ln \left(1+\frac{n}{x}\right)}-\frac{x}{\ln x}$$
$$=\frac{x\left(1+0\right)}{\ln x+\ln \left(1+0\right)}-\frac{x}{\ln x}$$
$$=\frac{x}{\ln x}-\frac{x}{\ln x}=0$$
the limit exists for all real n