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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.2012-06-16
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    The graph is [here](http://www.wolframalpha.com/input/?i=plot+%7Cx-3%7C)2012-06-16
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    Why would you down vote like that guys? Just comment, or upvote the comments that ask for improvement, or improve yourself.2012-06-16

1 Answers 1

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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?2012-06-16
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    In the the left sided limit I will use the number two to answer the question?2012-06-16
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    @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?2012-06-16
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    For the left sided limit replacing in the equation for 4 i got the -1 value.2012-06-16
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    @ViniciusL.Beserra Why would you replace by $4$? You should be evaluating the limit at $3^{-}$, not at $4$.2012-06-16
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    I used like a exaple from Leithold´s book.2012-06-17
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    @ViniciusL.Beserra Most probably, after simplifications, the author replaced $a$ in $\lim\limits_{x \to a}$. For example, if we want t find $\lim\limits_{x \to 7} 4x-3$ we would replace $7$ in the linear function and get $\lim\limits_{x \to 7} 4x-3=4\cdot 7-3=25$2012-06-17
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    In order to answer the question for a a left sided limit, i used u=x-3. I found 1 the for the left sided lmit.2012-06-17