Why Does this Limit as V approaches Infinity Equal Zero 
Given the following problem: $$\lim_{v\to\infty}\frac19(\ln|v-1|) - \frac19(\ln|v+8|)$$ 

Wouldn't this be undefined - as it would equal $\infty-\infty$?
However, Symbolab said it equaled zero.  
I'm solving a larger problem and this is just one of the last segments.  Please let me know why this would equal zero. Thank you. 
 A: Because you have
$$\frac{1}{9}\ln |v-1| - \frac{1}{9}\ln |v+8| = \frac{1}{9}\ln \frac{|v-1|}{|v+8|} $$ $$= \frac{1}{9}\ln \underbrace{\frac{|1-\frac{1}{v}|}{|1+\frac{8}{v}|}}_{\stackrel{v\to\pm\infty}{\longrightarrow}1}\stackrel{v\to \pm\infty}{\longrightarrow}\frac{1}{9}\ln 1 = 0$$
A: By the same argument, the limit $\lim_{x\to\infty}\bigl((x+1)-x\bigr)$ would be undefined (it is $\infty-\infty$ too), but it is actually equal to $1$.
Note that\begin{align}\log\lvert v+8\rvert&=\log\left\lvert(v-1)\frac{v+8}{v-1}\right\rvert\\&=\log\lvert v-1\rvert+\log\left\lvert\frac{v+8}{v-1}\right\rvert.\end{align}Can you take it from here?
A: Notice that
$$\frac19 \log |v-1| - \frac19 \log |v+8| = 
\frac19 \log \left| \frac{v-1}{v+8} \right|$$
and that
$$\lim_{v \to \infty} \frac{v-1}{v+8} = 1$$
Then
$$\lim_{v \to \infty} \frac19 \log \left| \frac{v-1}{v+8} \right| = \frac19 \log(1) = 0$$
since $\log$ is continuous.
A: As an aside, there is some ambiguity in your notation.  Some people would use $1/9(\ln|v-1|)$ to represent $\frac{1}{9 \ln |v-1|}$ and others to represent $\frac{1}{9} \ln |v-1|$.  Given you comments about the form of the limit, you seem to mean the latter.
"$\infty - \infty$" is one of several indeterminate forms.  We are responsible for recognizing indeterminate forms so that we can manipulate them to expose their limits.  (Two other indeterminate forms are "$\frac{0}{0}$", which turns up in every derivative, and "$\infty \cdot 0$", which turns up in every integral.)
For "$\infty - \infty$", try exponentiation.  They you are looking at $\frac{\mathrm{e}^\infty}{\mathrm{e}^\infty}$ for which there is some credible chance of cancellation.  Here, \begin{align*}
\lim_{v \rightarrow \infty} &\frac{1}{9} \ln |v-1| - \frac{1}{9} \ln |v+8|  \\
    &= \lim_{v \rightarrow \infty} \ln \left( \exp \left( \frac{1}{9} \ln |v-1| - \frac{1}{9} \ln |v+8| \right) \right)  \\
    &= \lim_{v \rightarrow \infty} \ln \left( 
        \frac{ \exp \left( \frac{1}{9} \ln |v-1| \right) }
        { \exp \left( \frac{1}{9} \ln |v+8| \right) } 
    \right)  \\
    &= \lim_{v \rightarrow \infty} \ln \left( 
        \frac{ \exp \left( \ln |v-1| \right)^{1/9} }
        { \exp \left( \ln |v+8| \right)^{1/9} } 
    \right)  \\
    &= \lim_{v \rightarrow \infty} \ln \left( 
        \frac{ |v-1|^{1/9} }{ |v+8|^{1/9} } 
    \right)  \\
    &= \lim_{v \rightarrow \infty} \ln \left( 
        \left( \frac{|v-1|}{|v+8|} \right)^{1/9} \right)  \\
    &= \lim_{v \rightarrow \infty} \frac{1}{9} \ln \left( \frac{|v-1|}{|v+8|} \right)  \\
    &= \frac{1}{9} \lim_{v \rightarrow \infty} \ln \left( \frac{|v-1|}{|v+8|} \right)  \text{.}
\end{align*}
We are taking a limit as $v \rightarrow \infty$, so we may assume that $v > 1$.  Therefore, the arguments to both absolute values are positive and we can replace them with their arguments. \begin{align*}
\lim_{v \rightarrow \infty} &\frac{1}{9} \ln |v-1| - \frac{1}{9} \ln |v+8|  \\
    &= \frac{1}{9} \lim_{v \rightarrow \infty} \ln \left( \frac{v-1}{v+8} \cdot \frac{1/v}{1/v}  \right)  \\
    &= \frac{1}{9} \lim_{v \rightarrow \infty} \ln \left( \frac{1-\frac{1}{v} }{1+\frac{8}{v}}  \right)  \\
    &= \frac{1}{9} \ln \lim_{v \rightarrow \infty} \left( \frac{1-\frac{1}{v} }{1+\frac{8}{v}}  \right)  & [\text{$\ln$ continuous near $1$}]  \\
    &= \frac{1}{9} \ln 1  \\
    &= \frac{1}{9} \cdot 0  \\
    &= 0  \text{.}
\end{align*}
A: As an alternative, recall that


*

*$x\to 0 \implies (1+ax)^\frac1x\to e^a$
then
$$\ln|v-1|) - \ln|v+8|=\\=\ln|v-1|)\color{red}{-\ln v} - \ln|v+8|\color{red}{+\ln v}=\\=\frac{1}v\left(\ln|1-1/v|^v) - \ln|1+8/v|^v\right) \to 0\cdot(1-8)\to 0$$
