Let y = f(x) = begin{cases} e^{-frac{1}{x^2}} | vedclass

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(C) First,we check the continuity at $x = 0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} e^{-\frac{1}{x^2}} = e^{-\infty} = 0$. Since $f(0) = 0$,the functi

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(C) First,we check the continuity at $x = 0$: $\lim_{x 	o 0} f(x) = \lim_{x 	o 0} e^{-\frac{1}{x^2}} = e^{-\infty} = 0$. Since $f(0) = 0$,the function is continuous at $x = 0$.Next,we check the differentiability at $x = 0$: $f'(0) = \lim_{h 	o 0} \frac{f(h) - f(0)}{h} = \lim_{h 	o 0} \frac{e^{-\frac{1}{h^2}}}{h}$. Let $t = \frac{1}{h}$,as $h 	o 0$,$t 	o \infty$. Then $\lim_{t 	o \infty} t e^{-t^2} = \lim_{t 	o \infty} \frac{t}{e^{t^2}} = \lim_{t 	o \infty} \frac{1}{2t e^{t^2}} = 0$. Thus,$f$ is differentiable at $x = 0$ with $f'(0) = 0$.As $x 	o \pm \infty$,$f(x) = e^{-\frac{1}{x^2}} 	o e^0 = 1$. The graph approaches the horizontal asymptote $y = 1$.Comparing the properties: the function is symmetric about the $y$-axis (even function),passes through $(0,0)$,has a horizontal tangent at $(0,0)$,and approaches $y = 1$ as $x 	o \pm \infty$. Option $C$ represents these features correctly.

Let $y = f(x) = \begin{cases} e^{-\frac{1}{x^2}}, & \text{if } x \neq 0 \\ 0, & \text{if } x = 0 \end{cases}$. Then which of the following can best represent the graph of $y = f(x)$?

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