Solving $\lim_{x\to+\infty }\frac{ \ln(x^2 - x +1)}{\ln(x^{10} + x +1)}$

Question

hints on solving $\lim_{x\to+\infty} \frac{ \ln(x^2 - x +1)}{\ln(x^{10} + x +1)}$ would be appreciated. i tried multiplying each polynomial with the inverse of highest power but it didnt work out.

no lhospital

Answer 1

As $x \to +\infty$, one may observe that $$ \frac{ \ln(x^2 - x +1)}{\ln(x^{10} + x +1)} = \frac{ \ln\left(x^2 (1-1/x+1/x^2)\right)}{ \ln\left(x^{10} (1+1/x^9+1/x^{10})\right)}\sim \frac{ \ln(x^2)}{\ln(x^{10})}=\frac{ 2\times\ln(x)}{10\times\ln(x)}=\frac15 $$

Answer 2

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