Mdthbs

As the limit approaches infinity of 1/9(ln|v-1|) - 1/9(ln|v+8|) wouldn't it equal undefined as it would be infinity minus infinity? However, Symbolab said it equaled zero. I'm solving a larger problem and this is just one of the last segments of it. Please let me know why this would equal zero. Thank you.

Because you have

$$frac{1}{9}ln |v-1| - frac{1}{9}ln |v+8| = frac{1}{9}ln frac{|v-1|}{|v+8|} $$ $$= frac{1}{9}ln underbrace{frac{|1-frac{1}{v}|}{|1+frac{8}{v}|}}_{stackrel{vtopminfty}{longrightarrow}1}stackrel{vto pminfty}{longrightarrow}frac{1}{9}ln 1 = 0$$

By the same argument, the limit $lim_{xtoinfty}bigl((x+1)-xbigr)$ would be undefined (it is $infty-infty$ too), but it is actually equal to $1$.

Note thatbegin{align}loglvert v+8rvert&=logleftlvert(v-1)frac{v+8}{v-1}rightrvert\&=loglvert v-1rvert+logleftlvertfrac{v+8}{v-1}rightrvert.end{align}Can you take it from here?

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.

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*}

Not really $lim_{x to infty} x = infty$, but you cannot say, that $lim_{x to infty} frac{x^2}{x}$ is undefined because it's division of two infinities. You can simplify your example:

$$lim_{x to infty} frac{1}{9}ln(x-1) - frac{1}{9}ln(x+8) = frac{1}{9} lim_{x to infty} ln(frac{x-1}{x+8}) = frac{1}{9} lim_{x to infty} ln(1 - frac{9}{x+8})=0$$

I have omited the absolute value, because if $x$ is large enough, those equations are equivalent.