Taylor theorem and Remainder term goes to $\infty$ I would follow the Lagrange version (Remainder term)
Let $f(x)$ be $n$th differientiable function and having a $(n+1)$th derivative
$f(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n+R_n(x)$ (Here the $R_n(x)$ is a remainder term)
$R_n(x) = {{f^{(n+1)}(t)} \over {(n+1)!}}(x-a)^{n+1}$ for some $t \in  (a,x) $ or $(x,a)$
So my question is "If the $x \to \pm\infty$, Does $t \to \pm \infty$"?
Thanks.
 A: Not necessarily. For example, the third-order Taylor polynomial with Lagrange remainder of the inverse tangent for positive values is
$$
\arctan x = x + \frac{3t^2  - 1}{3(1 + t^2 )^3}x^3 ,
$$
with a suitable $0<t<x$. Hence,
$$
0>\frac{\arctan x - x}{x^3 } = \frac{3t^2  - 1}{3(1 + t^2 )^3}.
$$
Since
$$
\mathop {\lim }\limits_{x \to  + \infty } \frac{\arctan x - x}{x^3 } = 0,
$$
and the rational function of $t$ is positive for $\frac{1}{\sqrt 3}<t$, we must have
$$
t \to \frac{1}{\sqrt 3}
$$
from the left as $x\to +\infty$ (and this $t$ is unique).
A: For instance, if the following conditions are met:


*

*$f$ is $C^\infty$ with compact support $K\subseteq[-M,M]$;

*$n>0$;

*$f^{(k)}(a)\ne 0$ for some $0<k\le n$;


then for large values of $x$ the function $f(x)-P_{a,n}(x)$ diverges. A fortiori, $R_{a,n}(x)$ must be non-zero eventually as $\lvert x\rvert\to\infty$ and therefore, for such $x$-s, any point $t_{n,a,x}$ such that $\frac{f^{(n+1)}(t_{n,a,x})}{(n+1)!}(x-a)^{n+1}=R_{a,n}(x)$ must be in $K$.
