I am obliged to find this limit:

$\rm\displaystyle\ \lim_{x\to\ - \infty }\ -\frac{1}{2}\ln \left|x-1 \right|+\frac{1}{4}\ln \left|x^{2}+x+1 \right|$

I have no idea how to start doing this. I am getting indeterminate form. I will be glad for any tips.

  • $\begingroup$ It would be better if you, at least, tried the problem! $\endgroup$ – NasuSama Jan 13 '14 at 21:47
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    $\begingroup$ Hint: reverse the two ln terms and bring the exponents inside the ln terms. Then combine into one ln term. Now try it! $\endgroup$ – imranfat Jan 13 '14 at 21:49

$$-\frac{1}{2}\ln \left|x-1 \right|+\frac{1}{4}\ln \left|x^{2}+x+1 \right|$$ $$=-\ln \left|x-1 \right|^{1/2} + \ln \left|x^{2}+x+1 \right|^{1/4}$$ $$=\ln \left| \frac{x^{2}+x+1}{(x-1)^2} \right|^{1/4}$$ What happens when $x\to\pm\infty?$ Can you see why the limit of the term inside the $\ln$ is $1$?

  • $\begingroup$ I see. I use L'Hôpital's rule twice and prove that $\rm\displaystyle\ \left| \frac{x^{2}+x+1}{(x-1)^2} \right| \rightarrow 1$ and then whole limit $\rightarrow 0$. My biggest concern was with bringing two logarithms into one. Why did you write : $\rm\displaystyle\ \ln \left| \frac{x^{2}+x+1}{(x-1)^2} \right|^{1/4}$ but not $\rm\displaystyle\ =\ln \left| \frac{{(x^{2}+x+1)}^{1/4}}{(x-1)^2} \right|$ ? $\endgroup$ – Marcin Majewski Jan 13 '14 at 21:58
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    $\begingroup$ That was unnecessary question. I figured it out already :D Thank you ! $\endgroup$ – Marcin Majewski Jan 13 '14 at 22:10

$-{1 \over 2} \log |1-x| + {1 \over 4} \log |x^2+x+1| = {1 \over 4} ( \log |x^2+x+1| -\log |x-1|^2 = {1 \over 4} \log | {x^2+x+1 \over x^2-2x +1 } | $.


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