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Let $f(x)=x{\times}(-1)^{\left \lfloor \frac1x \right \rfloor}$. Calculate its limit at $x=0$. According to me the limit doesn't exist because if I take log on both sides of the equation, I get:- $\ln f(x) = \ln x+\left \lfloor \frac1x \right \rfloor {\times} \ln(-1)$

Here $\ln(-1)$ doesn't exist and hence no limit should exist.

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    With positive numbers you can use logarithms freely. However, for limit questions, it is almost always a good idea if you *look* before starting to do algebraic manipulations.2012-06-24

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Note that $\left\lfloor \dfrac1x \right\rfloor$ denotes the greatest integer less than or equals $\dfrac1x$. Hence, $(-1)^{\left\lfloor \dfrac1x \right\rfloor}$ makes sense since the power is always an integer. All you need for this proof is that $(-1)^{\left\lfloor \dfrac1x \right\rfloor}$ is either $1$ or $-1$. Hence, we have that $-x \leq x \times (-1)^{\left\lfloor \dfrac1x \right\rfloor} \leq x$ Hence, as $x \to 0$, we have that $\lim_{x \to 0}-x \leq \lim_{x \to 0} x \times (-1)^{\left\lfloor \dfrac1x \right\rfloor} \leq \lim_{x \to 0} x$ Hence, $\lim_{x \to 0} x \times (-1)^{\left\lfloor \dfrac1x \right\rfloor} = 0$

EDIT

Note that $\log(a^b) = b \log(a)$ is valid only when $a>0$ and $x \in \mathbb{R}$. Hence, it is incorrect to write $\log((-1)^b) = b \log(-1)$.

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    @palio Yes. That is equivalent to what I have written.2012-06-24