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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calculus
absolute-value
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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