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.

Give an example of a function $f:\mathbb R \to \mathbb R$ that the limit $\displaystyle\lim_{x\to 0}(f(x)-f(2x))$ exists but $\displaystyle\lim_{x\to 0}f(x)$ does not exists.

I tried a few trig functions but they didn't work so any help would be appreciated.

share|improve this question

2 Answers 2

up vote 6 down vote accepted

How about $$f(x) = \begin{cases} 1,&x< 0\\ 2,&x \geq0.\end{cases}$$?

share|improve this answer
    
Doesn't this go to -1 or -2 when it's supposed to have a limit ? –  GinKin Dec 16 '13 at 17:39
1  
No; it becomes the $0$ function, since this function does not change under rescaling; if $x<0$, then $f(2x)=f(x)=1$, if $x\geq 0$ , then $f(2x)=f(x)=2$. –  user99680 Dec 16 '13 at 17:43

Try $$f(x) = \begin{cases}\ln |x|,&x\ne 0\\-\ln 2,&x= 0.\end{cases}$$

Edit $$ \lim_{x\to 0}(f(x)-f(2x) )= \lim_{x\to 0}(\ln |x|-\ln|2x|) = \lim_{x\to 0} (-\ln 2 ) =\ln 2.$$

On the other hand, $\lim_{x\to 0}\ln|x|=-\infty.$

share|improve this answer
    
I don't see how does the subtraction change the limit or it's absence. Can you explain a bit more please? –  GinKin Dec 16 '13 at 17:41

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.