Why Does this Limit as V approaches Infinity Equal Zero

Question

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.

Answer 1

Because you have

$$\frac{1}{9}\ln |v-1| - \frac{1}{9}\ln |v+8| = \frac{1}{9}\ln \frac{|v-1|}{|v+8|} $$ $$= \frac{1}{9}\ln \underbrace{\frac{|1-\frac{1}{v}|}{|1+\frac{8}{v}|}}_{\stackrel{v\to\pm\infty}{\longrightarrow}1}\stackrel{v\to \pm\infty}{\longrightarrow}\frac{1}{9}\ln 1 = 0$$

Answer 2

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?

Answer 3

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.

Answer 4

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*}

Answer 5

As an alternative, recall that

• $x\to 0 \implies (1+ax)^\frac1x\to e^a$

then

$$\ln|v-1|) - \ln|v+8|=\\=\ln|v-1|)\color{red}{-\ln v} - \ln|v+8|\color{red}{+\ln v}=\\=\frac{1}v\left(\ln|1-1/v|^v) - \ln|1+8/v|^v\right) \to 0\cdot(1-8)\to 0$$