Evaluating $\lim_{t \to 0} \left(\int_0^1[bx+a(1-x)]^t \mathrm dx \right)^{1/t}$ $0 \lt a \lt b,$ find 

$$\lim_{t \to 0} \left(\int_0^1[bx+a(1-x)]^t \mathrm dx \right)^{1/t}$$

I think substituting $(b-a)x+a=u$  will make
$$\begin{align*}
\lim_{t \to 0}\left(\frac1{b-a}\int_a^bu^t\ \mathrm du\right)^{1/t}
=&\,\lim_{t \to 0}\left(\frac{1}{b-a}\,\Bigl[\frac {u^{t+1}}{t+1}\Bigr]_a^b\right)^{1/t}\\ 
=&\,\left(\frac{1}{b-a}\left(\lim_{t \to 0} \frac{b^{t+1}}{t+1}-\frac{a^{t+1}}{t+1}\right)\right)^{1/t}
\end{align*}$$
This is where I got trouble, how can I evaluate this?
 A: Your solution was nice until you get
$$ L := \lim_{t\to 0} \left( \frac{1}{b-a} \frac{b^{t+1} - a^{t+1}}{t+1} \right)^{1/t}. $$
But you cannot put the limit inside as there is still the power $1/t$ to the outer parenthesis, which clearly depends on $t$. In this case, write
$$ b^{t} = e^{t\log b} = 1 + t \log b + \cdots $$
and likewise for $a^{t}$. Then
\begin{align*}
L
&= \lim_{t\to 0} \left( \frac{1}{b-a} \frac{b - a + (b\log b - a\log a)t + \cdots }{t+1} \right)^{1/t} \\
&= \lim_{t\to 0} \frac{1}{(1+t)^{1/t}} \left( 1 + \frac{b\log b - a\log a}{b-a} t + \cdots \right)^{1/t}.
\end{align*}
But we know that
$$ \lim_{t\to 0} (1 + xt)^{1/t} = e^{x}. $$
Therefore we have
$$ L = \exp\left( \frac{b\log b - a\log a}{b-a} - 1 \right). $$
A: $\lim_{t\to0}\bigg(\frac{1}{b-a}\int_{a}^{b}u^{t}dt\bigg)^{\frac{1}{t}}=e^{\lim_{t\to0}\frac{\ln\big(\frac{1}{b-a}\int_{a}^{b}u^{t}dt\big)}{t}}=e^{\lim_{t\to0}\frac{\ln\big(\frac{1}{(b-a)(t+1)}(b^{t+1}-a^{t+1})\big)}{t}}$
By L'hopital's rule we get that:
$\lim_{t\to0}\frac{\ln\big(\frac{1}{(b-a)(t+1)}(b^{t+1}-a^{t+1}\big)}{t}$
$=\lim_{t\to0}\frac{(b-a)(t+1)}{b^{t+1}-a^{t+1}}((b^{t+1}\ln(b)-a^{t+1}\ln(a))(b-a)(t+1))-(b^{t+1}-a^{t+1})(b-a))\frac{1}{(b-a)^{2}(t+1)^{2}}=\frac{b\ln(b)-a\ln(a)}{(b-a)}-1$.
Hence the limit to the original question is $e^{\frac{b\ln(b)-a\ln(a)}{b-a}-1}$.
A: As long as $0\lt a\le f(x)\le b\lt\infty$ on $[0,1]$,
$$
\lim_{t\to0}\frac1t\left(f(x)^t-1\right)=\log(f(x))
$$
converges uniformly on $[0,1]$. Therefore,
$$
\begin{align}
\lim_{t\to0}\left(\int_0^1f(x)^t\,\mathrm{d}x\right)^{1/t}
&=\lim_{t\to0}\left(1+t\int_0^1\frac1t\left(f(x)^t-1\right)\,\mathrm{d}x\right)^{1/t}\\
&=\lim_{t\to0}\left(1+t\int_0^1\log\left(f(x)\right)\,\mathrm{d}x\right)^{1/t}\\[6pt]
&=e^{\int_0^1\log\left(f(x)\right)\,\mathrm{d}x}
\end{align}
$$
