Calculus: Early Transcendentals 8th Edition

$$\lim\limits_{x\to\infty}[\ln(1+x^2)-\ln(1+x)]=\infty$$
$$A=\lim\limits_{x\to\infty}[\ln(1+x^2)-\ln(1+x)]$$$$A=\lim\limits_{x\to\infty}\Bigg[\ln\Big(\frac{1+x^2}{1+x}\Big)\Bigg]$$$$A=\ln\Bigg[\lim\limits_{x\to\infty}\Big(\frac{1+x^2}{1+x}\Big)\Bigg]$$ Divide both numerator and denominator by $x$, we have $$A=\ln\Bigg[\lim\limits_{x\to\infty}\Big(\frac{\frac{1}{x}+x}{\frac{1}{x}+1}\Big)\Bigg]$$ As $x\to\infty$, $(\frac{1}{x}+x)$ approaches $0+\infty=\infty$ while $(\frac{1}{x}+1)$ approaches $0+1=1$ Therefore, $\frac{\frac{1}{x}+x}{\frac{1}{x}+1}$ approaches $\infty$ So, $$A=\ln(\infty)=\infty$$