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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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    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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    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