# Proving a theorem on limits that approach infinity.

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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2022-07-25 20:46:07
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Not getting much into your general strategy to prove the said statement - if $-(\frac{x}{1})=-x>-(\frac{1}{N})^{\frac{1}{r}}=-(\frac{1}{N^{\frac{1}{r}}})$, than $-\frac{1}{x} < -(N)^{\frac{1}{r} }$.