Lim_(x=>0)(3^x-3^-x-2xlog_e3)/(x-sinx)=? 

প্রশ্ন: \( \lim_{x \to 0} \frac{3^x - 3^{-x} - 2x \log_e 3}{x - \sin x} = ? \)

সমাধান:

আমরা জানি, \( e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ... \)

সুতরাং, \( 3^x = e^{x \log_e 3} = 1 + x \log_e 3 + \frac{(x \log_e 3)^2}{2!} + \frac{(x \log_e 3)^3}{3!} + ... \)

এবং \( 3^{-x} = e^{-x \log_e 3} = 1 - x \log_e 3 + \frac{(x \log_e 3)^2}{2!} - \frac{(x \log_e 3)^3}{3!} + ... \)

তাহলে, \( 3^x - 3^{-x} = 2x \log_e 3 + 2 \frac{(x \log_e 3)^3}{3!} + 2 \frac{(x \log_e 3)^5}{5!} + ... \)

অতএব, \( 3^x - 3^{-x} - 2x \log_e 3 = 2 \frac{(x \log_e 3)^3}{3!} + 2 \frac{(x \log_e 3)^5}{5!} + ... \)

আমরা আরও জানি, \( \sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - ... \)

সুতরাং, \( x - \sin x = \frac{x^3}{3!} - \frac{x^5}{5!} + ... \)

এখন, \( \lim_{x \to 0} \frac{3^x - 3^{-x} - 2x \log_e 3}{x - \sin x} = \lim_{x \to 0} \frac{2 \frac{(x \log_e 3)^3}{3!} + 2 \frac{(x \log_e 3)^5}{5!} + ...}{\frac{x^3}{3!} - \frac{x^5}{5!} + ...} \)

\( = \lim_{x \to 0} \frac{2 \frac{x^3 (\log_e 3)^3}{3!} + O(x^5)}{\frac{x^3}{3!} + O(x^5)} \)

\( = \lim_{x \to 0} \frac{2 \frac{(\log_e 3)^3}{3!} + O(x^2)}{\frac{1}{3!} + O(x^2)} \)

\( = \frac{2 (\log_e 3)^3 / 3!}{1/3!} = 2 (\log_e 3)^3 \)

সুতরাং, \( \lim_{x \to 0} \frac{3^x - 3^{-x} - 2x \log_e 3}{x - \sin x} = 2 (\log_e 3)^3 \) 🥳