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Consider the continuity of (-1)^[1/x]


    \[ f(x) = (-1)^{\left \lfloor \frac{1}{x} \right \rfloor} \qquad \text{for } x \neq 0, \]

where \lfloor \ \rfloor denotes the greatest integer function, or floor function. Sketch the graph of f(x) for -2 \leq x \leq -\frac{1}{5} and \frac{1}{5} \leq x \leq 2. Evaluate

    \[ \lim_{x \to 0^+} f(x) \qquad \text{and} \qquad \lim_{x \to 0^-} f(x). \]

Is it possible to define f(0) in a way that makes f(x) continuous at 0.

First, we sketch the graph of f on the requested intervals.

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As x \to 0^+, f(x) alternates between +1 and -1.
As x \to 0^-, f(x) alternates between +1 and -1.

There is no way to define f(0) to make f continuous at 0 since f(x) will take both values +1 and -1 no matter how small we choose our \delta > 0. (So, if we were to try to define f(0) = 1, then for \varepsilon = \frac{1}{2} > 0 there is no \delta> 0 such that f(x) < \varepsilon whenever |x| < \delta, and similarly if we try to define f(0) = -1.)

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