- #1
IniquiTrance
- 190
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This might sound like a stupid question.
[itex] f(x) = \begin{cases} &e^{-\frac{1}{x^2}} &\text{if } x\neq 0 \\ & 0 &\text{if } x = 0 \end{cases}[/itex]
Is the reason f is infinitely differentiable at 0 because we keep differentiating 0 as a constant, or because,
[itex] \lim_{x\rightarrow 0} f`(x) = \lim_{x\rightarrow 0} \frac{2}{x^3} e^{-\frac{1}{x^2}} = 0[/itex]
Thanks.
[itex] f(x) = \begin{cases} &e^{-\frac{1}{x^2}} &\text{if } x\neq 0 \\ & 0 &\text{if } x = 0 \end{cases}[/itex]
Is the reason f is infinitely differentiable at 0 because we keep differentiating 0 as a constant, or because,
[itex] \lim_{x\rightarrow 0} f`(x) = \lim_{x\rightarrow 0} \frac{2}{x^3} e^{-\frac{1}{x^2}} = 0[/itex]
Thanks.