If lim _{t} {rightarrow 0}left(int_0^1(3 x+5) | vedclass

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(C) Let $L = \lim _{t ightarrow 0}\left(\int_0^1(3 x+5)^t d xight)^{\frac{1}{t}}$. This is of the form $1^{\infty}$.Using the formula $\lim_{t 	o

Vedclass · Accepted Answer

$64$

Vedclass · Answer

(C) Let $L = \lim _{t ightarrow 0}\left(\int_0^1(3 x+5)^t d xight)^{\frac{1}{t}}$. This is of the form $1^{\infty}$.Using the formula $\lim_{t 	o 0} f(t)^{g(t)} = e^{\lim_{t 	o 0} g(t)(f(t)-1)}$,we have:$L = e^{\lim_{t 	o 0} \frac{1}{t} \left( \int_0^1 (3x+5)^t dx - 1 ight)}$.Evaluating the integral: $\int_0^1 (3x+5)^t dx = \left[ \frac{(3x+5)^{t+1}}{3(t+1)} ight]_0^1 = \frac{8^{t+1} - 5^{t+1}}{3(t+1)}$.So,$L = e^{\lim_{t 	o 0} \frac{1}{t} \left( \frac{8^{t+1} - 5^{t+1}}{3(t+1)} - 1 ight)} = e^{\lim_{t 	o 0} \frac{8^{t+1} - 5^{t+1} - 3(t+1)}{3t(t+1)}}$.Applying $L$'Hopital's rule as $t 	o 0$: $\frac{8 \ln 8 - 5 \ln 5 - 3}{3}$.This simplifies to $e^{\ln(8^{8/3}) - \ln(5^{5/3}) - 1} = e^{\ln(\frac{8^{8/3}}{5^{5/3}}) - 1} = \frac{1}{e} \cdot \frac{8^{8/3}}{5^{5/3}} = \frac{1}{e} \cdot \frac{8^{2/3} \cdot 8^2}{5^{2/3} \cdot 5} = \frac{64}{5e} \left( \frac{8}{5} ight)^{2/3}$.Comparing with $\frac{\alpha}{5e} \left( \frac{8}{5} ight)^{2/3}$,we get $\alpha = 64$.

If $\lim _{t}$ ${\rightarrow 0}\left(\int_0^1(3 x+5)^t d x\right)^{\frac{1}{t}}=\frac{\alpha}{5 e}\left(\frac{8}{5}\right)^{\frac{2}{3}}$,then $\alpha$ is equal to . . . . . .

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