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I've been trying to do it and I still haven't been able to, I don't know what problem I have but it gives me 1, instead when I put it through some application it gives me the result of $ /1/2 * ln(2)$ I'm sorry if the result doesn't look good, I don't know how to use it. I am going to give you the process that I have done and there is the limit thank you very much.

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    $\begingroup$ $\ln p-\ln q= \ln(p/q)$ and not $\frac{\ln p}{\ln q}$ $\endgroup$ – DatBoi Oct 20 at 4:40
  • $\begingroup$ Are you talking about the first part? $\endgroup$ – Seis Oct 20 at 4:47
  • $\begingroup$ The very first step, yes $\endgroup$ – DatBoi Oct 20 at 4:51
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$$\lim_{x \to \infty} \ln((4x^3-4x^2+5)^{1/3})-\ln((2x^3-5x+4)^{1/3})$$

Now use that $\ln(a)-\ln(b)=\ln\left(\frac{a}{b}\right)$

$$=\lim_{x \to \infty} \ln\left(\frac{(4x^3-4x^2+5)^{1/3}}{(2x^3-5x+4)^{1/3}}\right)=\lim_{x \to \infty} \ln\left(\left(\frac{4x^3-4x^2+5}{2x^3-5x+4}\right)^{1/3}\right)$$

Next use $\ln(a^x)=x\cdot \ln(a)$

$$=\lim_{x \to \infty} \frac13\cdot \ln\left(\frac{4x^3-4x^2+5}{2x^3-5x+4}\right)$$

Finally expand the fraction by $\frac1{x^3}$ or you use L´Hospital. At this method you differentiate the numerator and the denominator three times.

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    $\begingroup$ I don't understand the last part, do I do it like I was doing in the third step? $\endgroup$ – Seis Oct 20 at 5:03
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    $\begingroup$ Do you know the rule $\ln(a^x)=x\ln(a)$? Or do you mean the expansion? $\endgroup$ – callculus Oct 20 at 5:05
  • $\begingroup$ no i think i haven't seen it $\endgroup$ – Seis Oct 20 at 5:07
  • $\begingroup$ This is one of the logarithm rules, for instance here. But there are many other sources. $\endgroup$ – callculus Oct 20 at 5:11
  • $\begingroup$ @Seis I think you need to review the basic properties of logarithms. You need to be comfortable with them if you are going to do calculus with logarithms. $\endgroup$ – PM 2Ring Oct 20 at 5:14
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As @DatBoi points out, we can get the result without needing to use L'Hopital. I get $\frac{\ln2}{3}$, since $$\lim_{x\rightarrow \infty}\frac{4x^3-4x^2+5}{2x^3-5x+4}\to2$$

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  • $\begingroup$ Yes, but I have to do the exercise by that method, that's the problem I have $\endgroup$ – Seis Oct 20 at 4:55
  • $\begingroup$ You could do the limit I indicated by using L'Hopital. The one within the logarithm. $\endgroup$ – Chris Custer Oct 20 at 4:56
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    $\begingroup$ Apply it three times, to get $24/12=2$. $\endgroup$ – Chris Custer Oct 20 at 4:57
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    $\begingroup$ Well, after the first application, you have $(12x^2-8x)/(6x^2-5)$. Numerator and denominator go to infinity again, so you can apply it again... $\endgroup$ – Chris Custer Oct 20 at 5:14
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    $\begingroup$ You might as well add this l'Hôpital stuff to your answer, rather than burying it down here in the comments. OTOH, I fully agree that using l'Hôpital for this limit is overkill, but hey, it's just a homework exercise... $\endgroup$ – PM 2Ring Oct 20 at 5:17

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