Mathematics - Limits Question with Solution | TestHub

$\begin{array}{rl}\frac{\alpha }{5e}& =\exp \left(\underset{t\to 0}{\lim }\frac{1}{t}\left({\int }_0^1(3x+5{)}^{t}dx-1\right)\right)\\ & =\exp \left(\underset{t\to 0}{\lim }\frac{1}{t}{\left(\frac{(3x+5{)}^{t+1}}{3(t+1)}\right)}_0^1-1\right)\\ & =\exp \left(\underset{t\to 0}{\lim }\frac{1}{t}\left(\frac{8^{t+1}-5^{t+1}}{3(t+1)}-1\right)\right)\\ & =\exp \left(\underset{t\to 0}{\lim }\frac{1}{t}\left(\frac{8^{t+1}-5^{t+1}-3t-3}{3(t+1)}\right)\right)\\ & =\exp \left(\underset{t\to 0}{\lim }\left(\frac{8^{t+1}\cdot \ln 8-5^{t+1}\ln 5-3}{3(t+1)}\right)\right)\\ & =\exp \left(\frac{\ln 8^8-\ln 5^5-3}{5}\right)\end{array}$ $\begin{array}{rl}& ={\left(\frac{8}{5}\right)}^{2/3}\frac{\alpha }{5e}=\exp \left(\frac{\ln \left(\frac{8^8}{5^5}\right)}{5}-1\right)\\ & \Rightarrow {\left(\frac{8}{5}\right)}^{2/3}\frac{\alpha }{5}={\left(\frac{8^8}{5^5}\right)}^{1/3}={\left(\frac{8^6\cdot 8^2}{5^3\cdot 5^2}\right)}^{1/3}=\frac{64}{5}{\left(\frac{8}{5}\right)}^{2/3}\\ & \Rightarrow \alpha =64\end{array}$