Let $0<a<b.$ Evaluate $$\lim_{t\to 0} \{ \int_{0}^{1} [bx+a(1-x)]^t dx\}^{1/t}.$$
[i][The final answer should not involve any other operations other than addition, subtraction, multiplication, division and exponentiation.][/i]
1. We start with the given limit:
   \[
   L = \lim_{t \to 0} \left\{ \int_{0}^{1} [bx + a(1-x)]^t \, dx \right\}^{1/t}
   \]

2. To simplify this, we use the fact that for a limit of the form \( \lim_{t \to 0} f(t)^{1/t} \), we can write:
   \[
   L = \exp \left( \lim_{t \to 0} \frac{\ln \left( \int_{0}^{1} [bx + a(1-x)]^t \, dx \right)}{t} \right)
   \]

3. Let:
   \[
   I(t) = \int_{0}^{1} [bx + a(1-x)]^t \, dx
   \]
   Then:
   \[
   L = \exp \left( \lim_{t \to 0} \frac{\ln I(t)}{t} \right)
   \]

4. We need to evaluate \( \lim_{t \to 0} \frac{\ln I(t)}{t} \). First, consider the integrand \( [bx + a(1-x)]^t \):
   \[
   [bx + a(1-x)]^t = \exp(t \ln(bx + a(1-x)))
   \]

5. Thus:
   \[
   I(t) = \int_{0}^{1} \exp(t \ln(bx + a(1-x))) \, dx
   \]

6. For small \( t \), we can approximate \( \exp(t \ln(bx + a(1-x))) \) using the first term of its Taylor series:
   \[
   \exp(t \ln(bx + a(1-x))) \approx 1 + t \ln(bx + a(1-x))
   \]

7. Therefore:
   \[
   I(t) \approx \int_{0}^{1} \left( 1 + t \ln(bx + a(1-x)) \right) \, dx
   \]

8. Splitting the integral:
   \[
   I(t) \approx \int_{0}^{1} 1 \, dx + t \int_{0}^{1} \ln(bx + a(1-x)) \, dx
   \]

9. Evaluating the first integral:
   \[
   \int_{0}^{1} 1 \, dx = 1
   \]

10. Let:
    \[
    J = \int_{0}^{1} \ln(bx + a(1-x)) \, dx
    \]

11. Thus:
    \[
    I(t) \approx 1 + tJ
    \]

12. Taking the natural logarithm:
    \[
    \ln I(t) \approx \ln(1 + tJ) \approx tJ \quad \text{(for small \( t \))}
    \]

13. Therefore:
    \[
    \lim_{t \to 0} \frac{\ln I(t)}{t} = \lim_{t \to 0} \frac{tJ}{t} = J
    \]

14. We need to evaluate \( J \):
    \[
    J = \int_{0}^{1} \ln(bx + a(1-x)) \, dx
    \]

15. Using the substitution \( u = bx + a(1-x) \), \( du = (b-a) \, dx \):
    \[
    J = \int_{a}^{b} \ln(u) \frac{du}{b-a}
    \]

16. This simplifies to:
    \[
    J = \frac{1}{b-a} \int_{a}^{b} \ln(u) \, du
    \]

17. The integral of \( \ln(u) \) is:
    \[
    \int \ln(u) \, du = u \ln(u) - u
    \]

18. Evaluating this from \( a \) to \( b \):
    \[
    \int_{a}^{b} \ln(u) \, du = \left[ u \ln(u) - u \right]_{a}^{b} = \left( b \ln(b) - b \right) - \left( a \ln(a) - a \right)
    \]

19. Therefore:
    \[
    J = \frac{1}{b-a} \left( (b \ln(b) - b) - (a \ln(a) - a) \right)
    \]

20. Simplifying:
    \[
    J = \frac{b \ln(b) - a \ln(a) - (b - a)}{b-a}
    \]

21. Finally, the limit is:
    \[
    L = \exp(J) = \exp \left( \frac{b \ln(b) - a \ln(a) - (b - a)}{b-a} \right)
    \]

The final answer is \(\boxed{\exp \left( \frac{b \ln(b) - a \ln(a) - (b - a)}{b-a} \right)}\)