Chapter 7: Transcendental Functions - Section 7.5 - Indeterminate Forms and L'Hopital's Rule - Exercises 7.5 - Page 409: 37

$\ln 2$

Here, we have $\lim\limits_{x \to \infty} f(\infty)=\lim\limits_{x \to \infty} \ln (\dfrac{2x}{x+1})=\dfrac{\infty}{\infty}$ This shows an indeterminate form of limit, thus we will apply L-Hospital's rule such as: $\lim\limits_{x \to \infty} f(x)=\lim\limits_{x \to \infty} \dfrac{p'(x)}{q'(x)}$ Now, $p'(x)=2$ and $q'(x)=1$ $ \lim\limits_{x \to \infty} \ln [\dfrac{2}{1}]=\ln 2$