2
$\begingroup$

How would i find the limit as $\lim\limits_{x\to3}\frac{4x(x-3)}{|x-3|}$? that is the absolute value of x-3 in the denominator. I thought my professor told my class that we were able to omit the absolute value sign for whatever reason. If that is true can you tell me why? Thanks!

  • 0
    @MichaelHardy: I still wouldn't say that this is a good advise. Think about $\operatorname{sgn} x$ with $a=0$.2011-09-28

2 Answers 2

4

You can find limits from each side. When taking the limit from the right, $x \gt 3$, so you can delete the absolute value signs. $\lim_{x \to 3^+}\frac{4x(x-3)}{|x-3|}=\lim_{x \to 3^+}\frac{4x(x-3)}{x-3}=\lim_{x \to 3^+}4x=12$ From the left, $x \lt 3$, so you must replace $|x-3|$ with $3-x$. $\lim_{x \to 3^-}\frac{4x(x-3)}{|x-3|}=\lim_{x \to 3^-}\frac{4x(x-3)}{3-x}=\lim_{x \to 3^-}-4x=-12$ As the left and right limits disagree, there is not a single limit.

  • 0
    I wish I hadn't asked that question after you had just shown that the function **tends to** -12 for x<3. I feel a bit stu$p$id. I hadn't seen WolframAlpha before joining this forum and I'm very impressed.2011-09-27
2

I guess you have to be told about the right and left limits: $ r = \lim\limits_{x\to3+0}\frac{4x(x-3)}{|x-3|} = \lim\limits_{x\to3+0}\frac{4x(x-3)}{x-3} = 12 $ $ l = \lim\limits_{x\to3-0}\frac{4x(x-3)}{|x-3|} = \lim\limits_{x\to3-0}\frac{4x(x-3)}{-(x-3)} = -12 $

The limit exists iff $r=l$ and in that case it is equal to each of them. That's not your case though.