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How does one get rid of the infinities arising here?

$$\lim_{x\to\infty}\left(\frac{\ln|x-1|}3-\frac{\ln|x+2|}3\right)$$

I really have no idea how to handle such natural logarithms.

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2 Answers 2

Hint: apply the rules of logarithm to get: $$\begin{align}\frac{\log|x-1|}{3} - \frac{\log|x+2|}{3} &= \frac{1}{3}(\log|x-1| - \log|x +2|) \\&= \frac{1}{3}\log\left|\frac{x-1}{x+2}\right|\\ &=\frac{1}{3}\log\left|\frac{1-\frac{1}{x}}{1 + \frac{2}{x}}\right|\end{align}$$

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HINT:

As $\displaystyle\ln a-\ln b=\ln (a/b)$

$$\ln|x-1|-\ln|x+2|=\ln\left|\frac{x-1}{x+2}\right|=\ln\left|\frac{1-1/x}{1+2/x}\right|$$

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