$displaystylelim_{xto 0}frac{1}{x}int_0^x e^{t^2} dt$

$F(0) = 0$, so the integral can be written as $$int_0^x e^{t^2} dt = F(x) - F(0).$$ Therefore, $$lim_{x to 0} frac{1}{x} int_0^x e^{t^2} dt = lim_{x to 0} frac{F(x) - F(0)}{x} = F'(0)$$ where the last equality follows directly from the definition of the derivative.

The integrand $e^{t^2}$ is continuous, so by the fundamental theorem of calculus, we have $$F'(x) = frac{d}{dx}int_0^x e^{t^2} dt = e^{x^2}$$ and from this it follows that $$F'(0) = e^{0^2} = e^0 = 1$$

$$L=lim_{x→0}frac{1}{x}int_0^x e^{t^2} dt=lim_{x→0}frac{int_0^x e^{t^2} dt}{x} left(frac 00 text{form}right)$$ So we have to use the L'Hospital rule. $$L=lim_{x→0}dfrac{frac d{dx}int_0^x e^{t^2} dt}{frac d{dx}x}=lim_{x→0}frac{ e^{x^2}}{1} =e^0=1.$$

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First fundamental theorem of calculus: Let $f$ be a continuous real-valued function defined on a closed interval $[a, b]$. Let $F$ be the function defined, for all $x$ in $[a, b]$, by

$$F(x)=int _{a}^{x}!f(t),dt~.$$ Then $F$ is uniformly continuous on $[a, b]$ and differentiable on the open interval $(a, b)$, and

$$F'(x)=dfrac{d}{dx}left(int _{a}^{x}!f(t),dtright)=f(x),$$ for all $x$ in $(a, b)~.$