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!
How do I find this limit?
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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
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.
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0I 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
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.