If $\lim_{t \to 0} \left(\int_0^1 (3x + 5)^t dx\right)^{\frac{1}{t}} = \frac{2}{56} \left(\frac{8}{5}\right)^{\frac{2}{\alpha}}$, then $\alpha$ is equal to:

Step 1: Evaluate the integral

First, we need to evaluate the definite integral $I(t) = \int_0^1 (3x + 5)^t dx$. We can use a substitution method for this integral. Let $u = 3x + 5$, then $du = 3dx$, so $dx = \frac{1}{3}du$. When $x=0$, $u=5$. When $x=1$, $u=8$.

Step 2: Substitute and integrate

Substitute $u$ and $du$ into the integral and evaluate it. The integral of $u^t$ with respect to $u$ is $\frac{u^{t+1}}{t+1}$ (for $t \neq -1$). Then, we apply the limits of integration from 5 to 8.

Step 3: Apply the limit using L'Hopital's Rule

The expression is in the indeterminate form $1^\infty$ as $t \to 0$. We can use the property that $\lim_{t \to 0} (f(t))^{1/t} = e^{\lim_{t \to 0} \frac{\ln(f(t))}{t}}$. Let $f(t) = \frac{8^{t+1} - 5^{t+1}}{3(t+1)}$. We need to evaluate the limit of the exponent using L'Hopital's Rule.

Step 4: Evaluate the limit of the exponent

Applying L'Hopital's Rule, we differentiate the numerator and the denominator with respect to $t$. The derivative of $\ln(g(t))$ is $\frac{g'(t)}{g(t)}$.

Step 5: Calculate derivatives and substitute $t=0$

We calculate the derivatives of the numerator and denominator. For the numerator, we use the chain rule and the derivative of $a^x$ is $a^x \ln a$. Then, we substitute $t=0$ into the resulting expression.

Step 6: Simplify the exponent and find the limit

We simplify the exponent using logarithm properties: $a \ln b = \ln b^a$ and $\ln a - \ln b = \ln(a/b)$. Then, we use $e^{\ln x} = x$ and $e^{-1} = 1/e$.

Step 7: Compare with the given expression to find alpha

We are given that the limit is equal to $\frac{2}{56} \left(\frac{8}{5}\right)^{\frac{2}{\alpha}}$. We need to equate our calculated limit to this expression and solve for $\alpha$. Note that $e \approx 2.718$. The given expression seems to have a typo, as $2/56 = 1/28$, and the $e$ term is missing. Assuming the problem intended to compare the powers of $8/5$, we focus on the exponent.

More from this chapter

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