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I want to prove the theorem $\lim_{x\to 0^-}\frac{1}{x^r}=+\infty$ if r is even. So that means I have to show that for any $N>0$ there exists a $\delta >0$ such that if $-x<\delta $ then $\frac{1}{x^r}>N$. First I solved for x in the 'then' statement so I got $x<(\frac{1}{N})^{1/r}$ then multiplied the inequality by -1 so $-x>-(\frac{1}{N})^{1/r}$ then this is the part which I might be wrong; I took the reciprocal of the right hand side of the inequality then I got: $-x<-N^{1/r}$. So if we can now take $\delta =-N^{1/r}$ and hence the theorem is proven?

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    Work with positive numbers. Situation is symmetric about $0$, so we only care about $|x|$ anyway.2012-05-13

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