if you are asking how $\displaystyle \lim_{x \to {\infty}}\frac{ln(x+1)}{ln(x)}=1$? it is either logical conisdering with huge numbers +1 doesnt matter or applying l'hopitals we get $\displaystyle \lim_{x \to {\infty}}\frac{\frac{1}{x+1}}{\frac{1}{x}}=\lim_{x \to {\infty}}\frac{x}{x+1}$ and either knowing the leading coefficent trick or applying l'hopital's again we get $\displaystyle \lim_{x \to {\infty}}\frac{1+0}{1}=1$