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