If lim t 0 ( 0 1 ( 3 x + 5 ) t d t ) 1 t = 5 e ( 8 5 ) 2 3 , then is equal to _________. [2025]

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Solution

Q.
If $\lim_{t \to 0} (\int_{0}^{1} {{(3 x + 5)}^{t} d t)}^{\frac{1}{t}} = \frac{α}{5 e} {(\frac{8}{5})}^{\frac{2}{3}}$ , then $α$ is equal to _________. [2025]

Ans.
(64)

Let $L = \lim_{t \to 0} (\int_{0}^{1} {{(3 x + 5)}^{t} d x)}^{\frac{1}{t}}$

$= e^{\lim_{t \to 0} \frac{1}{t} (\int_{0}^{1} {(3 x + 5)}^{t} d x - 1)}$

$= e^{\lim_{t \to 0} \frac{1}{t} [{(\frac{{(3 x + 5)}^{t + 1}}{3 (t + 1)})}_{0}^{1} - 1]}$ [ $∵ 1^{\infty} form$ ]

   $= e^{\lim_{t \to 0} [\frac{8^{t + 1} - 5^{t + 1} - 3 t - 3}{3 t (t + 1)}]}$

   $= e^{\lim_{t \to 0} {[\frac{8 \cdot 8^{t} - 5 \cdot 5^{t} - 3 t - (8 - 5)}{3 t}] \cdot \lim_{t \to 0} (\frac{1}{t + 1})}}$

$= e^{\lim_{t \to 0} \frac{1}{3} [\frac{8 (8^{t} - 1)}{t} - \frac{5 (5^{t} - 1)}{t} - \frac{3 t}{t}]}$

   $= e^{\frac{(8 \ln (8) - 5 \ln (5) - 3)}{3}} = e^{[\ln {(8)}^{8 / 3} - \ln {(5)}^{5 / 3} - \ln e]}$

$= \frac{{(8)}^{8 / 3}}{{(5)}^{5 / 3} \cdot e} \Rightarrow {(\frac{8}{5})}^{\frac{2}{3}} \cdot \frac{8^{2}}{5} \cdot \frac{1}{e} = \frac{α}{5 e} {(\frac{8}{5})}^{\frac{2}{3}}$ [Given]

On comparing, we get

$∴ α = 8^{2} = 64$ .

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