$1^{\infty}$ form

Now $\mathrm{L}=\mathrm{e}^{\mathrm{t} \rightarrow 0} \frac{1}{\mathrm{t}}\left(\left.\frac{(3 \mathrm{x}+5)^{\mathrm{t}+1}}{3(\mathrm{t}+1)}\right|_0 ^1-1\right)$

$$\begin{aligned} & =e^{t \rightarrow 0} \frac{8^{t+1}-5^{t+1}-3 t-3}{3 t(t+1)} \\ & =e \frac{8 \ln 8-5 \ln 5-3}{3} \\ & =\left(\frac{8}{5}\right)^{2 / 3}\left(\frac{64}{5}\right)=\frac{\alpha}{5 \mathrm{e}}\left(\frac{8}{5}\right)^{2 / 3} \end{aligned}$$

On comparing

$$\alpha=64$$