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. 

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*}
