what is the limit of: $\lim_{t \to 0}(\frac{1}{b-a}\int_{a}^{b}\left |f \right |^{t}(x)dx)^{\frac{1}{t}}$? edit:  we are using Riemann integration and f is continuous
given:
$f:[ a , b] \rightarrow \mathbb{R}$
$f(x) \neq 0$
i need to prove that:
$\lim_{t \to 0}(\frac{1}{b-a}\int_{a}^{b}\left |f  \right |^{t}(x)dx)^{\frac{1}{t}} = exp(\frac{1}{b-a}\int_{a}^{b}ln\left |f  \right |(x)dx)$
i got as fat as simplifing to: $\lim_{t \to 0}(1 +\frac{t}{b-a}\int_{a}^{b}ln\left |f  \right |(x)dx + O(t^{2}))^{\frac{1}{t}}$
i see it is colse to e but i cant understand how to get rid of the $O(t^{2})$
i tried to use the Squeeze theorem but i cant Squeeze the top to the e format

WLOG assume that $b-a=1$ (so we dont write $1/(b-a)$ all the time). So you have reached to 
$$\lim_{t\to0} (1+t\int_a^b\log|f|+O(t^2))^{1/t}\geq\lim_{t\to0}(1+t\int_a^b\log|f|)^{1/t}=\exp(\int_a^b\log|f|).$$
To get the other estimate, prove that for any $t>0$ that is sufficiently small we have the inequality $(1+tx)^{1/t}\leq e^x$ for all $x$. Let $M>0$ such that $O(t^2)\leq Mt^2$ in your notation. Then we have for any $t>0$, small,
$$(1+t\int_a^b\log|f|+O(t^2))^{1/t} \leq(1+t\int_a^b\log|f|+Mt^2)^{1/t}\leq\exp(\int_a^b\log|f|+Mt) $$
Taking limits as $t\to0$ yields $\lim_{t\to0} (1+t\int_a^b\log|f|+O(t^2))^{1/t}\leq\exp(\int_a^b\log|f|)$ as we wanted.
