Find $\lim_{x \to 0}\frac{\int_0^x(e^{2t}+t)^{1/t}dt}{x}$ It's asked to solve this:
$$\lim_{x \to 0}\frac{\int_0^x(e^{2t}+t)^{1/t}dt}{x}$$
And I have no idea how to do it...
 A: You can see this as the derivative of the function
$$F(x)=\int_0^x(e^{2t}+t)^{1/t}dt$$
At $x=0$ because $F(0)=0$
$$\lim_{x \to 0}\frac{\int_0^x(e^{2t}+t)^{1/t}dt-0}{x-0}=\lim_{x \to 0}\frac{F(x)-F(0)}{x-0}=F'(0)$$
Which by the fundamental theorem of calculus is equal to the function inside of the integral evaluated at $t=0$. But it happens that this value is not defined. So is needed to extend the function $(e^{2t}+t)^{\frac{1}{t}}$ continuously at $t=0$; this does not change the integral, and we can apply the FTC to the primitive of the extended function. 
To compute that continuous extension, we need to compute a new limit which is the value  of the function as $t \to 0$.
$$\lim_{t \to 0}(e^{2t}+t)^{1/t}=\exp{\lim_{t \to 0}\frac{\ln(e^{2t}+t)}{t}}$$
$$=\exp{\frac{d}{dt}\left(\ln{(e^{2t}+t)}\right)_{t=0}}=\exp{\left(\frac{2e^{2t}+1}{e^{2t}+t}\right)_{t=0}}=\exp{3}$$
A: L'Hopital's rule is only step 1:
$$\lim_{x \to 0}\frac{\int_0^x(e^{2t}+t)^{1/t}}{x}
=\lim_{x \to 0}\frac{ \frac{d}{dx} \left(\int_0^x(e^{2t}+t)^{1/t}\right) }{1} = 
\lim_{x \to 0}(e^{2x}+x)^{1/x}
$$
The next step is to expand $e^{2x}$:
$$
\lim_{x \to 0}(e^{2x}+x)^{1/x} = \lim_{x \to 0}\left([1+2x+O(x^2)] + x\right)^{1/x}  = \lim_{x \to 0}\left(1+3x\right)^{1/x} = e^3
$$
