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Please help me for solving $$\lim_{x\to\infty}\frac{\ln(x^2-x+1)}{\ln(x^{10}+x+1)}$$

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Have you seen L'Hôpital's rule? –  David Mitra Sep 11 '12 at 20:01
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You don't solve limits, you evaluate them. –  Stefan Smith Sep 16 '12 at 0:53

6 Answers 6

up vote 1 down vote accepted

$\lim_{x\to\infty}\frac{\ln(x^2-x+1)}{\ln(x^{10}+x+1)}$=$\lim_{x\to\infty}\frac{\ln x^2(1-\frac{1}{x}+\frac{1}{x^2})}{\ln x^{10} (1+\frac{1}{x^9}+\frac{1}{x^{10}})}$=$\lim_{x\to\infty}\frac{\ln x^2+\ln(1-\frac{1}{x}+\frac{1}{x^2})}{\ln x^{10}+\ln(1+\frac{1}{x^9}+\frac{1}{x^{10}})}$=$\lim_{x\to\infty}\frac{\ln x^2+\ln 1}{\ln x^{10}+\ln 1}$=$\lim_{x\to\infty}\frac{\ln x^2}{\ln x^{10}}$=$\lim_{x\to\infty}\frac{2\ln x}{10\ln x}$=$\frac{1}{5}$.

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Hint: For example the numerator can be written as $\ln(x^2-x+1)=\ln(x^2\cdot(1-\frac1x+\frac1{x^2}))= 2\ln(x) + \ln (1-\frac1x+\frac1{x^2})$. What can you say about $\ln (1-\frac1x+\frac1{x^2})$?

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You can use something called l'Hôpital's rule if both the numerator and the denominator of a fraction tend to $\infty$ or $0$. The rule says:

$$\lim \frac{f(x)}{h(x)}=\lim \frac{f'(x)}{h'(x)}$$

(provided the RHS exists)

In the example you gave:

$$\lim_{x\to\infty} \frac{\ln (x^2-x+1)}{\ln (x^{10}+x+1)}=\lim_{x\to\infty} \frac{\frac{2x-1}{x^2-x+1}}{\frac{10x^9+1}{x^{10}+x+1}}=\lim_{x\to\infty} \frac{2x^{11}\cdots}{10x^{11}\cdots}=\frac{1}{5}$$

(you could, of course, do without the rule in this case, as others have pointed out, but do keep it in mind as a possible resource when you encounter similar cases)

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All the answers here have a lot of calculation, and I find it surprising because the way I would have found the answer seems much simpler. I would say as follows: When $x$ is very large, the $x^2$ term dominates the behavior of $x^2-x+1$, and similarly $x^{10}$ dominates $x^{10}+x+1$. So for large $x$, the expression will behave like $$\frac{\ln x^2}{\ln x^{10}}.$$

This simplifies directly to $$2\ln x\over 10 \ln x$$ and the $\ln x$ terms cancel, leaving the correct answer, $\frac15$.

My sense is that this is the kind of reasoning used by experienced mathematicians and computer scientists. In this case it leads to the correct answer, and you can do it in your head in a few seconds. Are there cases where similar reasoning doesn't work? If not, why bother with the elaborate calculations given in the other answers?

(Addendum: an attempt to include more detail in this argument would yield something rather like Hagen von Eitzen's answer, but the point here is that even that level of detail is not really necessary, and perhaps that the more of it you can ignore, the better. Using $O$-notation would elide some of the detail from Hagen's answer; mine elides even more.)

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I'm sure many would do it this way, yet I highly doubt it whether any mathematician would teach or use this as a proof, much less in an exam or even homework. I, for one, did in my head to know the solution at once: $\,2/10\,$ , yet I'd follow one of the formal ways shown to write it down –  DonAntonio Sep 12 '12 at 2:05

Just use the dominant terms of the numerator and denominator and get: $$\lim_{x\to\infty}\frac{\ln(x^2-x+1)}{\ln(x^{10}+x+1)}=\lim_{x\to\infty}\frac{\ln(x^2)}{\ln(x^{10})}=\frac{1}{5}.$$

Q.E.D.

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Maybe I have an incorrect way of thinking about this, but because the x variable is all that changes you can see, as it approaches infinity, the most changing x for the denominator changes 5 times faster than the x of the numerator. From there you can just say 1/5.

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That is essentially, the idea behind all of the answers on this page. –  user39572 Sep 11 '12 at 20:47

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