Question

\mathop {Lim}\limits_{x \to 0} \frac{{{{16}^x} - {9^x}}}{{x({{16}^x} + {9^x})}} is equal to

• A. \ell n\left( {\frac{3}{2}} \right)
• B. \ell n\left( {\frac{2}{3}} \right)
• C. \ell n\left( {\frac{4}{3}} \right)
• D. ℓn2

Answer 1

Answer: (C)

\ell n\left( {\frac{4}{3}} \right)

Answer 2

\text{The given limit is } L = \mathop {Lim}\limits_{x \to 0} \frac{{{{16}^x} - {9^x}}}{{x({{16}^x} + {9^x})}}. \text{ Dividing the numerator and denominator by } 9^x, \text{ we get } L = \mathop {Lim}\limits_{x \to 0} \frac{{(\frac{16}{9})^x - 1}}{{x((\frac{16}{9})^x + 1)}}. \text{ This can be split into two limits: } \mathop {Lim}\limits_{x \to 0} \left( \frac{{(\frac{16}{9})^x - 1}}{x} \right) \times \mathop {Lim}\limits_{x \to 0} \left( \frac{1}{(\frac{16}{9})^x + 1} \right). \text{ Using the standard limit } \mathop {Lim}\limits_{y \to 0} \frac{{a^y - 1}}{y} = \ln a, \text{ the first limit is } \ln\left(\frac{16}{9}\right). \text{ The second limit evaluates to } \frac{1}{1+1} = \frac{1}{2}. \text{ Therefore, } L = \ln\left(\frac{16}{9}\right) \times \frac{1}{2} = \ln\left(\left(\frac{16}{9}\right)^{\frac{1}{2}}\right) = \ln\left(\frac{4}{3}\right).