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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!

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  • $\begingroup$ Maybe what the professor said was that if you're looking for $\lim_{x\to a}|f(x)|$, you can just find $\lim_{x\to a}f(x)$ without the absolute value, and then take an absolute value afterwards. That would make sense, and would clearly not apply to what you have here. $\endgroup$ – Michael Hardy Sep 27 '11 at 20:44
  • $\begingroup$ @MichaelHardy: I still wouldn't say that this is a good advise. Think about $\operatorname{sgn} x$ with $a=0$. $\endgroup$ – Ilya Sep 28 '11 at 8:14
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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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  • $\begingroup$ Am I right in thinking that this function is basically $y=4|x|$ but with a "hole" at $x=3$ where y takes a negative value? $\endgroup$ – Peter Phipps Sep 27 '11 at 20:18
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    $\begingroup$ No, you might take a look at wolframalpha.com/input/?i=plot+4x%28x-3%29%2Fabs%28x-3%29 It is $y=4x$ above $x \gt 3$, but is $y=-4x$ for $x \lt 3$ $\endgroup$ – Ross Millikan Sep 27 '11 at 20:20
  • $\begingroup$ There is a "hole" at $x=3$. A hole means $y$ does not have a value. For $x<3$, the function is $-4x$. This agrees with $4|x|$ when $x \le 0$, but not for $0<x<3$. For $x>3$ we are looking at $4x$, or equivalently $4|x|$. "Basically $4|x|$" is not right for $x$ in $(0,3)$. $\endgroup$ – André Nicolas Sep 27 '11 at 20:29
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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.

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