# 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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Work with positive numbers. Situation is symmetric about $0$, so we only care about $|x|$ anyway. – André Nicolas May 13 '12 at 18:37

What one must show is that, for every positive $N$, there exists a positive $\delta$ such that: $$\forall x\in\mathbb R,\quad-\delta\leqslant x\lt 0\implies 1/x^r\geqslant N.$$ So, fix $N\gt0$. Which $\delta\gt0$ would do the job?
You’re getting your signs a bit confused. The way you’ve set things up, your $x$ is negative, so it’s bound to be less than $\left(\frac1N\right)^{1/r}$; what’s important is that $-x<\left(\frac1N\right)^{1/r}$. Now you can multiply through by $-1$ to get $x>-\left(\frac1N\right)^{1/r}$, an inequality involving two negative numbers. When you take reciprocals, you must do it on both sides of the inequality. Since the two numbers have the same sign, you must also reverse the inequality: $$\frac1x<-N^{1/r}\;,$$ which isn’t very useful. But this last step was unnecessary. Go back and simply set $\delta=\left(\frac1N\right)^{1/r}$: then whenever $0<-x<\delta$ you have $$\frac1{-x}>\frac1\delta=N^{1/r}$$ and therefore $\dfrac1{(-x)^r}>N$. Finally, $r$ is even, so $(-x)^r=x^r$.
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 $-\frac{1}{x} < -(N)^{\frac{1}{r} }$.