it seems obvious that this integral is zero and so is the limit but what theorem we are using here?
I see it's connected to Riemann sums with an interval=zero Right ?
The function $\mathrm{f}$ is continuous.
$$\lim_{x \to 0}\int_0^x\mathrm{f}(x)\ \mathrm{d}x= \ ?$$
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$\begingroup$Assuming $f$ is Riemann integrable, it is bounded by some $B$, so $0 \le |\int_0^x f(x) dx | \le \int_0^x |f(x)| dx \le Bx$.
$\endgroup$ 5 $\begingroup$We are not using any theorem. The definition of the definite integral is $$ \int_a^b f(x) \; dx = \lim_{n\to \infty} \sum_{i=1}^n f(x_i)\Delta x. $$ where $x_i = a + i\Delta x$ and $\Delta x = \frac{b - a}{n}$. If $a=b=0$, then $\Delta x = 0$ and so the integral is zero: $$ \int_0^0 f(x)\; dx = \lim_{n\to \infty}\sum_{i=1}^n 0 = \lim_{n\to \infty} 0 = 0. $$
About the limit. Assume that $f$ is continuous on a small interval $[0, \epsilon]$. Then according to the Fundamental Theorem of Calculus the function given by $$ F(x) = \int_0^x f(t) \; dt $$ is continuous on $[0,\epsilon]$. In particular $F$ is continuous at $0$. This, by definition, means that $\lim_{x\to 0^+} F(x) = 0$, or that $$ \lim_{x\to 0^+} \int_0^x f(t) \; dt = 0. $$
$\endgroup$ 4 $\begingroup$Another approach valid when $f$ is continuos: the integral is a function $F$ s.t. $F′=f$ and $F(0)=0$. By continuity, $\lim F=F(0)=0$.
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