Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
You are using "laws of logarithms" where they do not apply. What is the exponent? –  André Nicolas Jun 24 '12 at 16:34
    
@AndréNicolas:- Please elaborate on your comment. Here $[\frac1x]$ is the exponent. So, I guess,I can use logarithm here. –  kusur Jun 24 '12 at 16:43
    
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. –  André Nicolas Jun 24 '12 at 16:47
    
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? –  kusur Jun 24 '12 at 16:54
    
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. –  André Nicolas Jun 24 '12 at 16:58

1 Answer 1

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 $\log((-1)^b) = b \log(-1)$.

share|improve this answer
    
When $x$ approaches 0, $\frac1x$ approaches infinity. Then what can we say about $[\frac1x]$ ? You are claiming this to be 1 or -1. Why? –  kusur Jun 24 '12 at 16:46
    
@KunalSuri $\left\lfloor \dfrac1x \right\rfloor$ also $\to \infty$. But what we need here is that it is always an integer as it tends to $\infty$. I am claiming that $(-1)^{\left\lfloor \dfrac1x \right\rfloor}$ is $1$ or $-1$. I am not claiming $\left \lfloor \dfrac1x \right \rfloor$ to be 1 or -1. I am only claiming that $\left \lfloor \dfrac1x \right \rfloor$ is an integer. –  user17762 Jun 24 '12 at 16:48
    
Thanks. Its just like $cosx$ when $x\to \infty$. It oscillates between 1 and -1. Here the Greatest integer function doesn't oscillate but still it can hold either of the two values - 1 or -1. right? –  kusur Jun 24 '12 at 16:52
    
@KunalSuri Note that $\left\lfloor \dfrac1x \right\rfloor$ doesn't oscillate. what oscillates is $(-1)^{\left\lfloor \dfrac1x \right\rfloor}$. It takes only values $-1$ and $1$. –  user17762 Jun 24 '12 at 16:55
    
Couldn't we just say that $|f(x)|=|x{\times}(-1)^{\left \lfloor \frac1x \right \rfloor}|=|x|\to 0$ then $f(x)=x{\times}(-1)^{\left \lfloor \frac1x \right \rfloor}$ also $\to 0$?. This is of course only valid when the limit is $0$. –  palio Jun 24 '12 at 17:55

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.