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.

I tried to do this exercise and i found the sided limits as 0, both left and right. But in the book the answer is -1 and 1. enter image description here

share|improve this question
Please, write that down in text, it is hard to read. –  Pedro Tamaroff Jun 16 '12 at 3:11
you took limits of $f(x)$, not $f'(x)$, that's why. –  Robert Mastragostino Jun 16 '12 at 3:17
The graph is here –  Santosh Linkha Jun 16 '12 at 3:25
Why would you down vote like that guys? Just comment, or upvote the comments that ask for improvement, or improve yourself. –  Pedro Tamaroff Jun 16 '12 at 4:45

1 Answer 1

If the function is

$$f(x) = |x-3|$$

then you need to find $f'_-(3)$ and $f'_+(3)$.

You can go as follows

$$f'_+(3)= \lim_{x \to 3^+} \frac{f(x)-f(3)}{x-3}$$

$$f'_+(3)= \lim_{x \to 3^+} \frac{|x-3|-0}{x-3}$$

since $x$ ranges over values greater than $3$, $|x-3|=x-3$, so

$$f'_+(3)= \lim_{x \to 3^+} \frac{x-3}{x-3}=1$$

Try to do it for $f'_-$, and note that since $x$ ranges over values smaller than $3$, $|x-3|=-(x-3)$.

share|improve this answer
Thanks Peter. In this case you used the number four to compute the limit? Am I right? –  Vinicius L. Beserra Jun 16 '12 at 3:51
In the the left sided limit I will use the number two to answer the question? –  Vinicius L. Beserra Jun 16 '12 at 3:54
@ViniciusL.Beserra Not really. Since the function $(x-3)/(x-3)$ is $=1$ at any point different from $3$, we conclude the limit as $x \to 3 $ is $1$. Do you follow? –  Pedro Tamaroff Jun 16 '12 at 4:47
For the left sided limit replacing in the equation for 4 i got the -1 value. –  Vinicius L. Beserra Jun 16 '12 at 23:21
@ViniciusL.Beserra Why would you replace by $4$? You should be evaluating the limit at $3^{-}$, not at $4$. –  Pedro Tamaroff Jun 16 '12 at 23:31

Your Answer


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.