# Problem on limit of Lagrange's form of remainder

For the function $$f(x) = \begin{cases} \ e^{\frac{-1}{x^2}} & x\neq0,\\\\ 0 & x=0 \end{cases}$$

Let $$R_n(x)$$ denote Lagrange's form of remainder. Is $$\lim_{n\to\infty} R_n(x)=0$$

Lagrange's form of remainder is $$R_n(x)=f(x)-\sum_{k=0}^{n}\frac1{k!}f^{(k)}(a)(x-a)^k$$.

But I am stuck on calculating the $$n+1^{th}$$ derivative of the give function and thereby applying limits.

• I don't understand what you mean. What's $a$ here? Is $limR_n(x)=0$ supposed to be $\lim_{x\to a} R_n(x)=0$ ? Is it $\lim_{n\to\infty} R_n(x)=0$ ? Pointwise? How is "Lagrange's form" relevant? $R_n(x)=f(x)-\sum_{k=0}^{n}\frac1{k!}f^{(k)}(a)(x-a)^k$ regardless of whether or not you observe that there is some $\xi(x)$ such that yada yada. And none of the limits I've mentioned depend on $\xi(x)$. Nov 16, 2023 at 12:23
• Lagrange's form is given in the question. We need to calculate it's limit pointwise. Sorry, I misunderstood the formula. I will make changes Nov 16, 2023 at 12:34
• What you've written is not Lagrange's form of the remainder. Nov 16, 2023 at 13:12
• I don't know much about Lagrange's remainder. I got these formulae from another site . Nov 16, 2023 at 18:39

What you've written as $$R_n(x)$$ is Tailor ramainder. The Lagrange form of the remainder comes from a theorem that ensures there is a point $$c$$ between $$a$$ and $$x$$ such that: $$R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1}$$ Then you shall prove that the expression at the right converges to $$0$$, which requires finding a general bound for the term $$f^{(n+1)}(c)$$.
However, the function you defined as $$f$$ is a classic example in analysis as a function that is $$C^n$$ but the Taylor polynomial centered at $$0$$ doesn't converges to $$f$$ at any point other than $$x=0$$. This is proved by checking that, for every $$n$$, $$f^{(n)}(x) = p_n(\frac{1}{x^2})e^{\frac{-1}{x^2}}$$, for some polynomial $$p_n$$.