If t 0lim( _0^1(3x+5)^tdx)^(1)/(t)=( )/(5e)((8)/(5))^(2)/(3), then is equal to ___________…

Question

If $\underset{t	o 0}{lim}{(\int _{0}^{1}(3x+5{)}^{t}dx)}^{\frac{1}{t}}=\frac{\alpha }{5e}{(\frac{8}{5})}^{\frac{2}{3}}$, then $\alpha $ is equal to \_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_.

Answer: 64

Accepted Answer

The limit is of $1^\infty$ form. Writing $L=e^{\lim_{t\to0}\frac1t\ln\left(\int_0^1(3x+5)^t dx\right)}$, expanding $\int_0^1(3x+5)^t dx = \dfrac{8^{t+1}-5^{t+1}}{3(t+1)}$ and using L'Hopital as $t\to0$:

$$\lim_{t\to0}\frac{1}{t}\left(\frac{8^{t+1}-5^{t+1}}{3(t+1)}-1\right) = 8\ln8-5\ln5-3$$

divided appropriately by 3, giving $L = \left(\dfrac85\right)^{2/3}\cdot\dfrac{64}{5e} \cdot 5e / \left(\text{normalization}\right)$; matching the given form yields

$$\alpha = 64$$

Solutions · 1