Question

If \(0<\mathrm{a}<\mathrm{b}\), find \(\lim _{\mathrm{t} \rightarrow 0}\left\\{_{0}^{1}\left[\int_{0}^{1 / \mathrm{b}} \mathrm{x}+\mathrm{a}(1-\mathrm{x})\right]^{\mathrm{t}} \mathrm{dx}\right\\}^{1 / \mathrm{t}}\)

Short answer

Question: Find the limit as \(t\) approaches 0 of the function: \(\lim _{t \rightarrow 0}\left\{_{0}^{1}\left[\int_{0}^{1 / b}[x + a(1 - x)]^{t} dx\right]^{\frac{1}{t}}\right\}\), where \(0 < a < b\). Answer: \(\frac{b}{ab-a^2}\)

Step-by-step solution

Simplify the Integral Expression

First, let’s simplify the integral expression. We have: \([x + a(1 - x)]^{t} = [a + (1 - a)x]^t\) Now, we need to integrate this expression with respect to \(x\) from 0 to \(1/b\): \(\int_{0}^{1/b}(a + (1 - a)x)^t dx\)

Apply the Substitution Rule

Let's apply the substitution rule (also known as u-substitution) to solve the integral. We set: \(u = a + (1 - a)x\) \(du = (1 - a)dx\) Rearranging to solve for \(dx\): \(dx = \frac{du}{1 - a}\) We also need to change the integration limits: When \(x = 0\), \(u = a + (1 - a) \cdot 0 = a\) When \(x = 1/b\), \(u = a + (1 - a) \cdot 1/b = a + (1 - a)/b\) Now, we can replace \(x\) with \(u\) and \(dx\) with its expression in terms of \(du\). After substitution, our integral becomes: \(\int_{a}^{a + (1-a)/b} \left(u^t \cdot \frac{du}{1 - a}\right)\)

Solve the Integral

Simplify and solve the integral: \(\frac{1}{1 - a}\int_{a}^{a + (1-a)/b} u^t du = \frac{1}{1 - a}\left[\frac{u^{t+1}}{t+1}\right]_{a}^{a + (1-a)/b}\) Now, substitute the limits of integration: \(\frac{1}{1 - a}\left[\frac{(a + (1-a)/b)^{t+1} - a^{t+1}}{t+1}\right]\)

Find the Limit

We now need to find \(\lim _{t \rightarrow 0}\left\{_{0}^{1}\left[\frac{(a + (1-a)/b)^{t+1} - a^{t+1}}{t+1}\right]^{\frac{1}{t}}\right\}\). To perform this limit, we can use L'Hopital's rule: \(\lim _{t \rightarrow 0}\left\{_{0}^{1}\left[\frac{(a + (1-a)/b)^{t+1} \ln(a + (1-a)/b) - a^{t+1} \ln(a)}{t+1}\right]^{\frac{1}{t}}\right\}\) Now we apply the exponent rule of limits and exponent laws: \(\lim _{t \rightarrow 0}\left\{_{0}^{1}\left[\frac{(a + (1-a)/b)^{t+1}}{(a)^{t+1}}\right]^{\frac{\ln(a + (1-a)/b) - \ln(a)}{t+1}}\right\}\) \(\lim _{t \rightarrow 0}\left\{_{0}^{1}\left[\frac{(1 + (b-a)/b)^{t+1}}{(a)^{t+1}}\right]^{\frac{\ln(a + (1-a)/b) - \ln(a)}{t+1}}\right\}\) Let's analyze the limit inside exponent. Using L'Hopital's rule again on \(\frac{\ln(a + (1-a)/b) - \ln(a)}{t+1}\), we have: \(\lim_{t\rightarrow0}\frac{\frac{1}{a+(1-a)/b}-\frac{1}{a}}{1} = \frac{1}{a} - \frac{b}{a+a(b-a)} = \frac{1}{a}-\frac{b}{ab}\) As \(t \rightarrow 0\), we have: \(\lim _{t \rightarrow 0}\left\{_{0}^{1}\left[\frac{(1 + (b-a)/b)^{t+1}}{(a)^{t+1}}\right]^{\frac{1}{a}-\frac{b}{ab}}\right\}\) = \([(1 + (b - a)/b)^{0+1}]/(a^{0+1}) = \frac{b}{ab-a^2}\) Finally, we have the result for the given limit: \(\lim _{t \rightarrow 0}\left\{_{0}^{1}\left[\int_{0}^{1 / b}[x + a(1 - x)]^{t} dx\right]^{\frac{1}{t}}\right\} = \frac{b}{ab-a^2}\)