Using L'Hopital to find $\lim_{x→∞} ( \ln \sqrt[3]{4x^3-4x^2+5} - \ln \sqrt[3]{2x^3-5x+4} )$

Question

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.

Answer 1

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

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.

Answer 2

As @DatBoi points out, we can get the result without needing to use L'Hopital. I get $\frac{\ln2}{3}$, since $$\lim_{x\rightarrow \infty}\frac{4x^3-4x^2+5}{2x^3-5x+4}\to2$$