If f(x) = frac{x - e^x + cos 2x}{x^2} for x n | vedclass

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(D) For $f(x)$ to be continuous at $x = 0$,$f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{x - e^x + \cos 2x}{x^2} = \lim_{x 	o 0} \frac{x - (1 +

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$[f(0)] \cdot \{f(0)\} = -1.5$

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(D) For $f(x)$ to be continuous at $x = 0$,$f(0) = \lim_{x 	o 0} f(x)$.$\lim_{x 	o 0} \frac{x - e^x + \cos 2x}{x^2} = \lim_{x 	o 0} \frac{x - (1 + x + \frac{x^2}{2!} + \dots) + (1 - \frac{(2x)^2}{2!} + \dots)}{x^2}$$= \lim_{x 	o 0} \frac{x - 1 - x - \frac{x^2}{2} + 1 - 2x^2}{x^2} = \lim_{x 	o 0} \frac{-\frac{5}{2}x^2}{x^2} = -\frac{5}{2} = -2.5$.Thus,$f(0) = -2.5$.Now,$[f(0)] = [-2.5] = -3$.And $\{f(0)\} = f(0) - [f(0)] = -2.5 - (-3) = 0.5$.Therefore,$[f(0)] \cdot \{f(0)\} = (-3) \cdot (0.5) = -1.5$.

If $f(x) = \frac{x - e^x + \cos 2x}{x^2}$ for $x \neq 0$ is continuous at $x = 0$,then which of the following is true? (Note: $[x]$ and $\{x\}$ denote the greatest integer and fractional part functions,respectively.)

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