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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0Maybe 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. – 2011-09-27
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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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0Am 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? – 2011-09-27
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1No, you might take a look at http://www.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 $x \lt 3$ – 2011-09-27
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0There 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 $0
3$ we are looking at $4x$, or equivalently $4|x|$. "Basically $4|x|$" is not right for $x$ in $(0,3)$. – 2011-09-27 -
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 stupid. 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.