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I tried to do the exercise below and I found the one-sided limits as 0, both left and right. But in the book the answer is -1 and 1.

Make the graph of the function. Determine if the function is continuous at $c$. Compute the lateral limits $f_-'(x_1)$ and $f_+'(x_1)$. $f(x)=|x-3|$; $x_1=3$.

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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 $3$, $|x-3|=-(x-3)$.

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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

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