Why is this function continuous over $x$?
$$f^{n}(\xi(x))=\frac{-(n!\left(-f(x)+f(x_{0})+f(x_0)(x-x_0)+\ldots+\frac{f^{n-1}(x_0)*(x-x0)^{n-1}}{(n-1)!}\right)}{(x-x_0)^n}$$
I tried solving it via the Heine sequence criteria, but to no avail. I need someone help me with the explanation for this, please!
Could the answer be, that since the left side of the equation is the final remainder of a Taylor series and a Taylor series converges to a certain function, then that remainder would ideally converge to zero?
Use the Taylor formula with the integral form of the remainder term $$ f(x)=f(x_0)+f'(x_0)(x-x_0)+\ldots+\frac{f^{(n-1)}(x_0)(x-x_0)^{n-1}}{(n-1)!} \\ +\int_{x_0}^x\frac{f^{(n)}(s)(x-s)^{n-1}}{(n-1)!}ds. $$ Now substitute $s=x_0+t(x-x_0)$ in the remainder term to get $$ \int_{x_0}^x\frac{f^{(n)}(s)(x-s)^{n-1}}{(n-1)!}ds=(x-x_0)^n\int_0^1\frac{f^{(n)}(x_0+t(x-x_0))(1-t)^{n-1}}{(n-1)!}dt $$ to see that indeed the quotient is a continuous function as long as $f$ is $n$ times continuously differentiable.