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Single Variable Calculus

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Convergence of Taylor Series to the Original Function

SVCALC-4W9I1U

Consider the function:

$$f(x)=\begin{cases} e^{-1/x}\quad x>0 \\\ 0\quad\quad x\leq0\end{cases}$$

It can be shown that $f(x)$ is infinitely differentiable. Let $a$ be a positive number.

Does the Taylor series of $f(x)$ at $x=0$ converge for all $x$ values in the interval $(-a,a)$, and does $f(x)$ equal its Taylor series on $(-a,a)$?

(The answer to this question doesn't depend on the exact value of $a$. To answer this question, start by computing the derivatives of $f(x)$ at $0$.)

A

The Taylor series of $f(x)$ at $x=0$ converges on $(-a,a)$, and $f(x)$ equals its Taylor series on $(-a,a)$.

B

The Taylor series of $f(x)$ at $x=0$ converges on $(-a,a)$, but $f(x)$ doesn't equal its Taylor series on $(-a,a)$.

C

The Taylor series of $f(x)$ at $x=0$ doesn't converge on $(-a,a)$, and $f(x)$ equals its Taylor series on $(-a,a)$.

D

The Taylor series of $f(x)$ at $x=0$ doesn't converge on $(-a,a)$, and $f(x)$ doesn't equal its Taylor series on $(-a,a)$.