Need clarification on this l'Hopital's rule solution I found a solution to this limit using l'hoptital's rule, but I don't get the third step.
$$\eqalign{ \lim_{x\to\infty} {\ln3x}-{\ln(x+1)}
&= \lim_{x\to\infty} \ln\frac{3x}{x+1} \\
&=\lim_{x\to\infty} \ln\frac{3}{1+1/x}\\
&=\ln3\\}$$
I'm not sure how what happened to get the third line. Can you explain?
 A: Without L'Hospital's Rule
First note that 
$$ \frac{a}{b}=\frac{ac}{bc}\quad\forall (a,b,c)\in\mathbb{R}:bc\ne 0$$
Since $x\to\infty$ then clearly $x\gt 0$, so let $c=\frac{1}{x}$
$$ \lim_{x\to\infty} \ln\left(\frac{3x}{x+1}\right) = \ln\left(\lim_{x\to\infty} \frac{3x}{x+1}\right)= \ln\left(\lim_{x\to\infty} \frac{3}{1+\frac{1}{x}}\right)= \ln\left(\frac{3}{1+0}\right)=\ln(3) $$
With L'Hospital's Rule
I wouldn't recommend L'Hospital's rule for this limit, however here it is
$$ \lim_{x\to\infty} \ln\left(\frac{3x}{x+1}\right) = \ln\left(\lim_{x\to\infty} \frac{\frac{d}{dx}[3x]}{\frac{d}{dx}[x+1]}\right)=
\ln\left(\lim_{x\to\infty} \frac{3}{1+0}\right)= \ln\left(\frac{3}{1+0}\right)=\ln(3) $$
A: $$\frac{3x}{x+1}=\frac{3\cdot x}{(1+\frac{1}{x})\cdot x}=\frac{3}{1+\frac{1}{x}}\cdot\frac{\not x}{\not x}=\frac{3}{1+\frac{1}{x}}$$
A: Divide both the numerator and the denominator by $x$ (which leave the fraction $\frac{3x}{x+1}$ unchanged)
A: If you are saying you only problem is step 3, it is a short answer:
It is established that the limit of f(x)= 1/x as x approaches positive infinity, is 0. Thus, the result is the log of 3/(1+0) = 3/1, or, simply, ln 3.  
