Answer
$$\frac{1}{5}$$
Work Step by Step
$$\eqalign{
& \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {{e^x} + 1} \right)}}{{5x}} \cr
& {\text{Verify if the conditions of l'Hospital's rule are satisfied}}{\text{. Here}} \cr
& {\text{evaluate the limit substituting }}\infty {\text{ for }}x{\text{ into the numerator and }} \cr
& {\text{denominator}}{\text{.}} \cr
& \mathop {\lim }\limits_{x \to \infty } = \ln \left( {{e^0} + 1} \right) = \ln \left( \infty \right) = \infty \cr
& \mathop {\lim }\limits_{x \to \infty } 5x = 5\left( \infty \right) = \infty \cr
& {\text{Since the limits of both numerator and denominator are }}\infty {\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. then differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {{e^x} + 1} \right)}}{{{D_x}\left( {5x} \right)}} \cr
& {D_x}\left( {\ln \left( {{e^x} + 1} \right)} \right) = \frac{1}{{{e^x} + 1}}\left( {{e^x}} \right) = \frac{{{e^x}}}{{{e^x} + 1}} \cr
& {D_x}\left( {5x} \right) = 5 \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{\frac{{{e^x}}}{{{e^x} + 1}}}}{5} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}{{5\left( {{e^x} + 1} \right)}} \cr
& {\text{Find the limit of the quotient of the derivatives}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}{{5\left( {{e^x} + 1} \right)}} = \frac{{{e^\infty }}}{{5\left( {{e^\infty } + 1} \right)}} = \frac{\infty }{\infty } \cr
& {\text{Since the limits of both numerator and denominator are }}\infty {\text{, }} \cr
& {\text{l'Hospital's rule applies}}{\text{. then differentiating the numerator and}} \cr
& {\text{denominator}} \cr
& = \mathop {\lim }\limits_{x \to \infty } \frac{{{D_x}\left( {{e^x}} \right)}}{{{D_x}\left( {5\left( {{e^x} + 1} \right)} \right)}} = \mathop {\lim }\limits_{x \to \infty } \frac{{{e^x}}}{{5{e^x}}} = \mathop {\lim }\limits_{x \to \infty } \frac{1}{5} \cr
& {\text{Find the limit }} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{1}{5} = \frac{1}{5} \cr
& {\text{By l'Hospital' rule}}{\text{, this result is the desired limit:}} \cr
& \mathop {\lim }\limits_{x \to \infty } \frac{{\ln \left( {{e^x} + 1} \right)}}{{5x}} = \frac{1}{5} \cr} $$
