$\lim\limits_{x\to\infty}\frac{\log x}{\log 7x}$? I did l'Hospital's rule and I got 
$\dfrac{1}{x}/\dfrac{7}{7x} = \dfrac{0}{0}$ (after "plugging" $\infty$ in)
But is that the answer? I dont think $\frac{0}{0}$ is an answer. Or is it the answer and that means the limit is divergent.
 A: To properly exploit L'hopital's rule, one must know the limit exists; see this wiki article.  Here we kill two birds with one stone, showing that there is a limit and that it is $1$, without using L'hopital or derivatives:
For $x > 1$, $\ln x > 0$, whence we can say:
$\dfrac{\ln x}{\ln 7x} = \dfrac{\ln x}{\ln x + \ln7} = \dfrac{1}{1 + \dfrac{\ln7}{\ln x}} \to 1 \tag{1}$
as $\ln x \to \infty$.  As $x \to \infty$, $\ln x \to \infty$, and so . . . 
voila!, followeth (1)!
A: $$\lim_{x\to \infty}\frac{\ln x}{\ln7x}$$
Let $f(x)=\dfrac{\ln x}{\ln7x}$ as you it goes to in determinant, so L'Hopital's rule gives $$\frac{\frac{1}{x}}{\frac{7}{7x}}=\frac{7x}{7x}=1$$
i.e., the limit value is 1
A: Something else to note, using property of logs, we have $\log (7x)=\log(7) +\log (x)$.   So your fraction can be rewritten as 
$\frac {\log x}{\log x +\log 7}$.  Since $\log 7$ is constant,  as $x\to \infty$, we have $\log x \to \infty$,  so the tiny increase is meaningless, thus a limit of 1.   Formally, you'd still probably use L'Hospitals as above.
A: Don't be hasty to put the x value too soon:
$$\require{cancel}\lim_{x\to\infty}\frac{\ln x}{\ln 7x}\stackrel{\rm L'Hos.}\lim_{x\to\infty}\frac{\frac1{\cancel{\color{blue}x}}}{\frac{1}{\cancel{\color{red}7}\cancel{\color{blue}x}}.\cancel{\color{red}7}}=1$$
where I have used the chain rule for log7x.
