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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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    You are using "laws of logarithms" where they do not apply. What is the exponent?2012-06-24
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    @AndréNicolas:- Please elaborate on your comment. Here $[\frac1x]$ is the exponent. So, I guess,I can use logarithm here.2012-06-24
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    Don't use logarithms, the logarithm is often undefined, unless you go to complex numbers, and even there $\log$ behaves weirdly. Look directly at your function. The reason I think your exponent $[1/x]$ must be the greatest integer $\le 1/x$ is that for general real $y$, $(-1)^y$ is undefined.2012-06-24
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    so what you suggested was just to avoid confusion due to the presence of a negatuve number, right? This means that if there was some other postive number in place of -1, then I could have used logarithm?2012-06-24
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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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