Question

\(\rm \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}} {{\tan \left( {5x} \right)}}\times\frac{{\left( {1 - \cos 4x} \right)\left( {{5^x} - {4^x}} \right)}}{{{x^3}}}} \right)\) equals to
1. \(\frac{8}{3}\ell n\left( {\frac{5}{2}} \right)\)
2. \(\frac{8}{5}\ell n\left( {\frac{5}{2}} \right)\)
3. \(\frac{8}{5}\ell n\left( {\frac{5}{4}} \right)\)
4. \(\frac{8}{3}\ell n\left( {\frac{5}{4}} \right)\)

Answer 1

Correct Answer - Option 3 : \(\frac{8}{5}\ell n\left( {\frac{5}{4}} \right)\)

Concept:

Formulas:

1 - cos 2θ = 2 sin2 θ

1 + cos 2θ = 2 cos2 θ

\(\rm \mathop {{\rm{lim}}}\limits_{x \to 0} \;\;\frac{{\sin x}}{x} = 1\)

\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{\tan x}}{x}} = 1\)

\(\rm \mathop {\lim }\limits_{x\; \to 0} {\frac{{a^x-1}}{x}} = \ln a\)

 

 

Calculation:

\(\rm \mathop {\lim }\limits_{x \to 0} \left( {\frac{{\sin x}} {{\tan \left( {5x} \right)}}\times\frac{{\left( {1 - \cos 4x} \right)\left( {{5^x} - {4^x}} \right)}}{{{x^3}}}} \right)\)

= \(\rm \mathop {\lim }\limits_{x \to 0} {\frac{{\sin x}} {{\tan \left( {5x} \right)}}\times \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos 4x} \right)\left( {{5^x} - {4^x}} \right)}}{{{x^3}}}} \)

= \(\rm \mathop {\lim }\limits_{x \to 0} {\frac{{\sin x}} {{\tan \left( {5x} \right)}}\times \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \cos 4x} \right)}}{{{x^2}}}} \times \mathop {\lim }\limits_{x \to 0} \frac {\left( {{5^x} - {4^x}} \right)}{x} \)

= \(\rm \mathop {\lim }\limits_{x \to 0} {\frac{{\sin x}} {{x}} \times \mathop {\lim }\limits_{x \to 0} \frac {5x}{\tan 5x} \times \frac 1 5\times \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2\sin^2 2x} \right)}}{{{x^2}}}} \times \mathop {\lim }\limits_{x \to 0} \frac {\left( {{5^x - 1)} - {(4^x-1}} \right)}{x} \)

= \(\rm \mathop {\lim }\limits_{x \to 0} {\frac{{\sin x}} {{x}} \times \mathop {\lim }\limits_{x \to 0} \frac {5x}{\tan 5x} \times \frac 1 5\times \mathop {\lim }\limits_{x \to 0} \frac{{\left( {2\sin^2 2x} \right)}}{{{x^2}}}} \times \mathop {\lim }\limits_{x \to 0} \frac {\left( {{5^x - 1)} - {(4^x-1}} \right)}{x} \)

= \(\rm \mathop {\lim }\limits_{x \to 0} {\frac{{\sin x}} {{x}} \times \mathop {\lim }\limits_{x \to 0} \frac {5x}{\tan 5x} \times \frac 1 5\times \mathop {\lim }\limits_{x \to 0} \frac{{\left( {8\sin^2 2x} \right)}}{{{(2x)^2}}}} \times \mathop {\lim }\limits_{x \to 0} \frac {\left( {{5^x - 1)} - {(4^x-1}} \right)}{x} \)

= \(\rm \frac 85 \times \left[ \mathop {\lim }\limits_{x \to 0} \frac {\left( {{5^x - 1)} } \right)}{x} - \mathop {\lim }\limits_{x \to 0} \frac {\left( {{4^x - 1)} } \right)}{x} \right]\)

\(= \frac{8}{5}\left( {\ln 5 - \ln 4} \right) = \frac{8}{5}\ln \left( {\frac{5}{4}} \right)\)