"x^2 / (x-3)" how can we simplfy this by making some cancellations so that we can get limit for 3? thanks...
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Hello, Originally Posted by zorspas "x^2 / (x-3)" how can we simplfy this by making some cancellations so that we can get limit for 3? thanks... Hmmm if you want the limit when x tends to 3, you don't need to simplify... $\displaystyle \lim_{x \to 3} x^2=9$ $\displaystyle \lim_{x \to 3^+} \frac{1}{x-3}=+ \infty$ $\displaystyle \lim_{x \to 3^-} \frac{1}{x-3}=- \infty$
Since right hand limit =+infinity and left hand limit =-infinity (already shown by Moo) therefor limit at x=3 does not exist.
though there is no need for simplification, what can be done to simplify it? thanks...
Originally Posted by zorspas though there is no need for simplification, what can be done to simplify it? thanks... $\displaystyle \frac{x^2}{x-3} = \frac{[x^2 - 9] + 9}{x-3} = \frac{(x-3)(x+3) + 9}{x-3} = \frac{(x-3)(x+3)}{x-3} + \frac{9}{x-3} = x+3 + \frac{9}{x-3}$. Or you can do polynomial long division and get the same result.
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