Let $g(x)= \begin{cases} e^{-1/x^2} & x\neq0 \\ 0 & x=0 \\ \end{cases}$
Additionally, show that $g(x)$ is differentiable for all $x\in\mathbb R$ with $g'(0)=0$.
Show that $g(x)$ is continuous at 0.
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calculus
real-analysis
proof-writing
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0Do you know what it means for a function to be continuous at a point? – 2017-02-16
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0This is a well known example of a function that is infinitely differentiable at the origin where all derivatives vanish. As such, the Taylor Series converges nowhere but at $x=0$ and is constant regardless of how many terms of the Taylor Series we take, even though our original function is smooth, not a second degree polynomial, and non-zero – 2017-02-16
1 Answers
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HINT
To show it's continuous at zero you need to show that $$\lim_{x\to 0}g(x) = \lim_{x\to 0 } e^{-1/x^2}=g(0) = 0.$$
To show it's differentiable at zero you must show that $$\lim_{x\to 0} \frac{g(x)-g(0)}{x} = \lim_{x\to 0} \frac{e^{-1/x^2}}{x}$$ exists. (And that it equals $0$ in order to show $g'(0) = 0.$)