Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2 - Page 426: 55

$$\eqalign{ & \left( {\text{a}} \right)3 \cr & \left( {\text{b}} \right) - 5 \cr} $$

$$\eqalign{ & \left( {\text{a}} \right)\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 3x} \right)}}{x} \cr & {\text{Rewrite the limit}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 3x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 3x} \right) - \ln \left( {1 + 3\left( 0 \right)} \right)}}{{x - 0}} \cr & {\text{The derivative of the function }}f\left( x \right){\text{ at the point }}x = a,{\text{ is}} \cr & f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}},{\text{ then}} \cr & \underbrace {\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 3x} \right) - \ln \left( {1 + 3\left( 0 \right)} \right)}}{{x - 0}}}_{\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \Rightarrow f\left( x \right) = \ln \left( {1 + 3x} \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {1 + 3x} \right)} \right] = \frac{3}{{1 + 3x}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 3x} \right) - \ln \left( {1 + 3\left( 0 \right)} \right)}}{{x - 0}} = f'\left( 0 \right) \cr & f'\left( 0 \right) = \frac{3}{{1 + 3\left( 0 \right)}} = 3 \cr & {\text{Therefore,}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 + 3x} \right)}}{x} = 3 \cr & \cr & \left( {\text{b}} \right)\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 - 5x} \right)}}{x} \cr & {\text{Rewrite the limit}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 - 5x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 - 5x} \right) - \ln \left( {1 - 5\left( 0 \right)} \right)}}{{x - 0}} \cr & {\text{The derivative of the function }}f\left( x \right){\text{ at the point }}x = a,{\text{ is}} \cr & f'\left( a \right) = \mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}},{\text{ then}} \cr & \underbrace {\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 - 5x} \right) - \ln \left( {1 - 5\left( 0 \right)} \right)}}{{x - 0}}}_{\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \Rightarrow f\left( x \right) = \ln \left( {1 - 5x} \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {1 - 5x} \right)} \right] = - \frac{5}{{1 - 5x}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 - 5x} \right) - \ln \left( {1 - 5\left( 0 \right)} \right)}}{{x - 0}} = f'\left( 0 \right) \cr & f'\left( 0 \right) = - \frac{5}{{1 - 5\left( 0 \right)}} = - 5 \cr & {\text{Therefore,}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {1 - 5x} \right)}}{x} = - 5 \cr} $$