How to find $\lim_{t\to0}\bigg(\int_0^1[bx+a(1-x)]^tdx\bigg)^{1/t}$ If $0\lt a\lt b$, I'm trying to find:
$$\lim_{t\to0}\bigg(\int_0^1[bx+a(1-x)]^tdx\bigg)^{1/t}$$
Solving the integral by substitution we have:
$$\lim_{t\to 0}\bigg(\frac{1}{b-a}\cdot\frac{b^{t+1}-a^{t+1}}{t+1}\bigg)^{1/t}$$
I don't how to proceed. 
What I know:


*

*This identity
$b^{t+1}-a^{t+1}=(b-a)(b^t+b^{t-1}a+\ldots+ba^{t-1}+a^t)$

*This limit is in the form: $1^{\infty}$.


I think I'm close, I need a hand how to find this limit.
 A: Apply l'Hospital's rule to the limit of the logarithm of the given integral. This leads to:
$$
\lim_{t \to 0} \left( \frac{\log( \int_{0}^{1} [bx + a(1-x)]^t \mathrm dx ) }{t} \right) = \lim_{t \to 0} \left( \frac{\int_{0}^{1} [bx + a(1-x)]^t \log [bx + a(1-x)] \mathrm dx}{\int_{0}^{1} [bx + a(1-x)]^t \mathrm dx} \right) = \int\limits_{0}^{1} \log [bx + a(1-x)] \mathrm dx = \frac{1}{b-a} (\log \frac{b^b}{a^a})-1\,.
$$
Above, use of Leibniz's integral rule was made. Consequently, the desired limit is 

$$
e^{-1}\left(\frac{b^b}{a^a}\right)^{1/(b-a)}
$$  

A: suppose we make a change of variable $y = bx + a(1-x),$ then we have $$\begin{align}\int_0^1[bx+a(1-x)]^tdx &= \frac1{b-a}\int_a^by^t \, dt\\
&=\frac{b^{t+1} - a^{t+1}}{(b-a)(t+1)}
\end{align}$$
now we also have 
$$\begin{align}\lim_{t\to 0}\left( \frac{b^{t+1} - a^{t+1}}{b-a}\right)^{1/t} &= \lim_{t\to 0} \left(\frac{e^{(1+t)\ln b} - e^{(1+t)\ln a}}{b-a}\right)^{1/t}\\
 &=\left(\frac{b(1+t\ln b+\cdots)-a(1 + t\ln a+\cdots)}{b-a}\right)^{1/t}\\
&=\left(1 +t(b\ln b - a\ln a + \cdots)\right)^{1/t}\\
&=e^{(b\ln b - a \ln a)/(b-a)}\\
\lim_{t \to 0}(1 + t)^{1/t} &= e\end{align}$$ 
therefore $$ \lim_{t\to0}\left(\int_0^1[bx+a(1-x)]^tdx\right)^{1/t} = e^{-1+(b\ln b - a \ln a)/(b-a)}.$$
A: Being sloppy whenever
I feel like it:
$\begin{array}\\
\bigg(\int_0^1[bx+a(1-x)]^tdx\bigg)^{1/t}
&=\bigg(\int_0^1[a+x(b-a)]^tdx\bigg)^{1/t}\\
&=\bigg(\int_0^1[a+cx]^tdx\bigg)^{1/t}
\qquad\text{where }c = b-a > 0\\
&=\bigg(\frac1{c}\int_a^bu^tdu\bigg)^{1/t}
\qquad u = a+cx, dx = du/c\\
&=\bigg(\frac1{c}\frac{b^{t+1}-a^{t+1}}{t+1}\bigg)^{1/t}
\qquad\text{as user42912 got}\\
&\approx\bigg(\frac1{c}\frac{b^{1+1/n}-a^{1+1/n}}{1+1/n}\bigg)^{n}
\qquad t = 1/n\\
&=\bigg(\frac1{c}\frac{b\,b^{1/n}-a\,a^{1/n}}{1+1/n}\bigg)^{n}\\
&\approx\bigg(\frac1{c}\frac{b(1+\ln b/n)-a(1+\ln a/n)}{1+1/n}\bigg)^{n}\\
&\approx\bigg(\frac1{c}\frac{b-a+(b\ln b-a\ln a)/n}{1+1/n}\bigg)^{n}\\
&\approx\bigg(\frac{b-a}{c(1+1/n)}+\frac1{c}\frac{(b\ln b-a\ln a)/n}{1+1/n}\bigg)^{n}\\
&=\bigg(\frac{n}{n+1}+\frac1{c}\frac{b\ln b-a\ln a}{n+1}\bigg)^{n}\\
&=\bigg(1-\frac1{n+1}+\frac1{c}\frac{b\ln b-a\ln a}{n+1}\bigg)^{n}\\
&=\bigg(1+\frac1{c}\frac{b\ln b-a\ln a-c}{n+1}\bigg)^{n}\\
&\approx\exp\bigg(\frac{b\ln b-a\ln a-c}{b-a}\bigg)
\qquad\text{since }(1+x/n)^n \approx e^x\\
&\approx\exp\bigg(\frac{b\ln b-a\ln a}{b-a}-1\bigg)\\
&=\frac1{e}\bigg(\frac{b^b}{a^a}\bigg)^{\frac1{b-a}}
\end{array}
$
And that's what I get.
As a check,
if $a=0$ and $b=1$,
the expression is
$\begin{array}\\
\left(\int_0^1 x^t dt\right)^{1/t}
&=\left(\frac{x^{t+1}}{t+1}|_0^1\right)^{1/t}\\
&=\left(\frac{1}{t+1}\right)^{1/t}\\
&\to \frac1{e}
\qquad\text{since } (1+t)^{1/t} \to e\\
\end{array}
$
