# 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.

• the $t$ is not unique, so the question does not really makes sense. Mar 2, 2020 at 7:34
• @GreginGre, In my thought Since $t$ are not unique, Could we make the sequence of the $t$? If so the $x$ goes to infinity for example, then there are sequence of the $t$ goes to infinity. Mar 2, 2020 at 8:37

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 $$0. Hence, $$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}, we must have $$t \to \frac{1}{\sqrt 3}$$ from the left as $$x\to +\infty$$ (and this $$t$$ is unique).
1. $$f$$ is $$C^\infty$$ with compact support $$K\subseteq[-M,M]$$;
2. $$n>0$$;
3. $$f^{(k)}(a)\ne 0$$ for some $$0;
then for large values of $$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$$.