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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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    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.2011-09-27
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    @MichaelHardy: I still wouldn't say that this is a good advise. Think about $\operatorname{sgn} x$ with $a=0$.2011-09-28

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