Chapter 6 - Exponential, Logarithmic, And Inverse Trigonometric Functions - 6.2 Derivatives And Integrals Involving Logarithmic Functions - Exercises Set 6.2 - Page 426: 57

$$\eqalign{ & \left( {\text{a}} \right)0 \cr & \left( {\text{b}} \right)\sqrt 2 \cr} $$

$$\eqalign{ & \left( {\text{a}} \right)\mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {\cos x} \right)}}{x} \cr & {\text{Rewrite the limit}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {\cos x} \right)}}{x} = \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {\cos x} \right) - \ln \left( {\cos 0} \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( {\cos x} \right) - \ln \left( {\cos 0} \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( {\cos x} \right) \cr & f'\left( x \right) = \frac{d}{{dx}}\left[ {\ln \left( {\cos x} \right)} \right] = \frac{{ - \sin x}}{{\cos x}} = - \tan x \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {\cos x} \right) - \ln \left( {\cos 0} \right)}}{{x - 0}} = f'\left( 0 \right) \cr & f'\left( 0 \right) = - \tan \left( 0 \right) = 0 \cr & {\text{Therefore,}} \cr & \mathop {\lim }\limits_{x \to 0} \frac{{\ln \left( {\cos x} \right)}}{x} = 0 \cr & \cr & \left( {\text{b}} \right)\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{\sqrt 2 }} - 1}}{h} \cr & {\text{Rewrite the limit}} \cr & \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{\sqrt 2 }} - 1}}{h} = \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{\sqrt 2 }} - {{\left( {1 + 0} \right)}^{\sqrt 2 }}}}{{h - 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_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{\sqrt 2 }} - {{\left( {1 + 0} \right)}^{\sqrt 2 }}}}{{h - 0}}}_{\mathop {\lim }\limits_{x \to a} \frac{{f\left( x \right) - f\left( a \right)}}{{x - a}}} \Rightarrow f\left( h \right) = {\left( {1 + h} \right)^{\sqrt 2 }} \cr & f'\left( h \right) = \frac{d}{{dh}}\left[ {{{\left( {1 + h} \right)}^{\sqrt 2 }}} \right] = \sqrt 2 {\left( {1 + h} \right)^{\sqrt 2 - 1}} \cr & \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{\sqrt 2 }} - {{\left( {1 + 0} \right)}^{\sqrt 2 }}}}{{h - 0}} = f'\left( 0 \right) \cr & f'\left( 0 \right) = \sqrt 2 {\left( {1 + 0} \right)^{\sqrt 2 - 1}} = \sqrt 2 {\left( 1 \right)^{\sqrt 2 - 1}} \cr & f'\left( 0 \right) = \sqrt 2 \cr & {\text{Therefore,}} \cr & \mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {1 + h} \right)}^{\sqrt 2 }} - 1}}{h} = \sqrt 2 \cr} $$