Find the sided limits

Question

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 $f_+'(x_1)$. $f(x)=|x-3|$; $x_1=3$.

Why would you down vote like that guys? Just comment, or upvote the comments that ask for improvement, or improve yourself. — Pedro, Jun 16 '12 at 04:45

score 6 · Answer 1 · answered Jun 16 '12 at 03:15

6

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

answered Jun 16 '12 at 03:15

Pedro

122,002

Thanks Peter. In this case you used the number four to compute the limit? Am I right? – Vinicius L. Beserra Jun 16 '12 at 03:51
In the the left sided limit I will use the number two to answer the question? – Vinicius L. Beserra Jun 16 '12 at 03:54
1

@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 Jun 16 '12 at 04:47
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 Jun 16 '12 at 23:31
I used like a exaple from Leithold´s book. – Vinicius L. Beserra Jun 17 '12 at 01:41
@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 $\lim\limits_{x \to 7} 4x-3=4\cdot 7-3=25$ – Pedro Jun 17 '12 at 01:43
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. – Vinicius L. Beserra Jun 17 '12 at 11:38

Find the sided limits

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