Why is $\lim_{x \to \infty}\ln\left(\frac{x}{x+1}\right) = 0$? This equation goes like this: $$\lim_{x \to \infty}\ln\left(\frac{x}{x+1}\right) = 0$$
Can you help me understand the logic behind the above equation?
 A: The logarithm function is continuous (so that the limit of the logarithm is the logarithm of the limit) and its argument tends to $1$ (because the inverse, $1+1/x$, tends to $1$).
A: $$\lim_{x\to\infty}\dfrac{x}{x+1}=1$$
$$\ln(1)=0$$
A: substitute $u\mapsto \frac{x}{x+1}$ now you're left with
$$ \lim_{u\to 1} \ln u = \ln 1 = 0 $$ 
In English: As $x$ gets bigger, the fraction $\frac{x}{x+1}$ gets closer and closer to $1$. Write out several terms to see this: $\frac 3 4$, $\frac 4 5$, $\ldots$, $\frac{999}{1000}$, $\ldots$, $\frac{999\ 999}{1\ 000\ 000} \to 1$. And we know $\ln(1) = 0$.
A: See the Limits of compositions of functions theorem
$\left\lbrace\begin{array}l \lim \limits_{x\to +\infty} \dfrac{x}{x+1} =1 \\\\ \lim \limits_{X\to 1} \ln X =0 \end{array}\right.\implies \lim \limits_{x\to +\infty} \ln\left(\dfrac{x}{x+1}\right) =0$
A: $$\lim_{x \to \infty}\ln \left(\frac{x}{x+1}\right)=\ln \left(\lim_{x \to \infty}\frac{x}{x+1}\right)=\ln \left(\lim_{x \to \infty}\frac{1}{1+\frac1x}\right)=\ln 1 = 0$$
The first equality is due to continuity of $\ln$ function.
