For example, say I have a function $f(x)$ defined from $-1$ to $1$ inclusive and it is Riemann integrable. Then, I have a sequence defined as $$\int_{\frac{-1}{n}}^{\frac{1}{n}} f(x) dx = x_n$$
How would I then evaluate the limit of $x_n $?
And would there be a corresponding proof of this necessary with the computation?
Thank you for all the help!
On
Let $$I_n=\int_{-\frac{1}{n}}^{+\frac{1}{n}} f(x)\, dx$$ Develop the integrand as a Taylor series around $x=0$ $$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)$$ $$\int f(x)\,dx=f(0) x+\frac{1}{2} x^2 f'(0)+\frac{1}{6} x^3 f''(0)+O\left(x^4\right)$$ Apply the bounds to get $$I_n=\frac{2 f(0)}{n}+\frac{f''(0)}{3 n^3}+O\left(\frac{1}{n^5}\right)$$
Since $f$ is Riemann integrable, $f(x)$ is bounded. Let $\sup$ of |f| be $M$. So $|x_n| \le \frac{2M}{n}$. So given $p>0$, there exists $N>0$ such that $|x_n| \le \frac{2M}{n} < p$ for all $n\ge N$, because of the $n$ in the denominator. So $\lim x_n= 0$.