Prove that $\lim_{x\to -\infty}x^n=\infty$ for $n\in \Bbb N$ and $n$ even, and $\lim_{x\to -\infty}x^n=-\infty$ for $n\in \Bbb N$ and $n$ odd.
I tried solving this as follows:
If $n$ is odd, then we try to show that, $\lim_{x\to -\infty}x^n=-\infty$ for $n\in \Bbb N.$ In other words, we use the following definition that:
Let $f:A\to\Bbb R$, where $A\subseteq \Bbb R$ and let $\exists a\in\Bbb R$ such that $(-\infty,a)\subseteq A.$ If $\alpha\in \Bbb R$ then there exist a $K<a$ such that for all $x<K$ we have, $f(x)\lt\alpha$ then, $\lim_{x\to -\infty}f(x)=-\infty$ and vice-versa.
So, if we can show that the conditions in the definition above, holds true, when $f(x)=x^n$ and $A=\Bbb R$ we are done. We may take, $a$ as any arbitrary real number as, the domain of $f$ is continuous.
If $\alpha\in\Bbb R$ be any arbitrary real number, then, we fix, $K=\inf\{ \alpha, -1\}$ and note that, if $x\lt K\leq -1$ then, $x\lt -1$ and so, $x^2\gt 1\implies x^{2k+1}\lt x\lt K\leq \alpha,$ $k\in\Bbb N.$ This shows that $x^{2k+1}=x^{n}\lt \alpha$ where, $n$ is an odd natural number for all, $x\lt K.$ This proves that $\lim_{x\to -\infty}x^n=-\infty$ where, $n$ is an odd natural number from the definition above.
Next, I tried show that $\lim_{x\to -\infty}x^n=\infty$ for $n\in \Bbb N$ and $n$ even. I went on as:
We use the below definition:
Let $f:A\to\Bbb R$, where $A\subseteq \Bbb R$ and let $\exists a\in\Bbb R$ such that $(-\infty,a)\subseteq A.$ If $\alpha\in \Bbb R$ then there exist a $K<a$ such that for all $x<K$ we have, $f(x)\gt\alpha$ then, $\lim_{x\to -\infty}f(x)=\infty$ and vice-versa.
So, if we can show that the conditions in the definition above, holds true, when $f(x)=x^n$ and $A=\Bbb R$ we are done. We may take, $a$ as any arbitrary real number as, the domain of $f$ is continuous.
Let $\alpha\in \Bbb R$ be an arbitrary real number and we fix, $K=\sup\{\alpha,1\}$ and consider $x\lt -K.$ Then, $-x\gt K\geq 1$ and hence, $x^2\gt -x\gt K\geq \alpha$ and $x^{2k}\geq x^2$ (because, $x^2\gt K\geq 1$) implies, $x^{2k}\gt \alpha,$ where $k$ is a positive integer. So, if $n=2k$ then, $x^n\gt\alpha.$
So, we have just shown that $\forall \alpha\in\Bbb R$ $,\exists K=\sup\{\alpha,1\}$ such that $\forall x\lt -K$ we have, $x^n\gt\alpha$ for all even positive integer $n.$ This means, $\lim _{x\to -\infty}x^n=\infty,$ when $n$ is an even positive integer.
Is this a valid way to solve this problem? Is there any alternative/simpler way to solve these sort of problems?
Finally, can the proof-writing be improved? Any suggestions are welcome. Last but not the least, is that, if there are any errors please do let me know.
P.S. : It's the first time I am working with the definitions of "Infinite Limits at Infinity."
Let $n\in \Bbb N$. Set $f(x):=x^{2n}, x\in \Bbb R$. Let $M>0$; we need a $c>0$ such that $x^{2n}\geq M$, if $x\leq-c$. We observe that $x^{2n}\geq M\iff x\geq M^{\frac {1}{2n}}$, so we choose $c:=M^{\frac {1}{2n}}$ and we have $x\leq -M^{\frac {1}{2n}}\implies -x\geq M^{\frac {1}{2n}}\implies |x|\geq M^{\frac {1}{2n}}\implies x^{2n}\geq M$.
The other limit is quite similar (for arbitrary $M>0$ you need to find similarly a suitable $c>0$).