proof that k times a limit at negative inifinity equals negative infinity.

Question

below is a proof I am stuck on- I am uncertain as to where to go past the final step (If I've even done it correctly). Any help would be greatly appreciated. Thanks

Answer 1

$$\lim_{x\to\infty} g(x)=-\infty$$ means for any preassingned real number $N$ there exists a $m\in \mathbb{R}$ such that $g(x) \leq (-N)$ whenever $x>m$. Now choose $N'=\frac{N}{k}$ ,$k>0$. Then there exists $m'\in \mathbb{R}$ such that $g(x) \leq (-N')$ whenever $x>m'$

i.e, $g(x) \leq (-\frac{N}{k})$ whenever $x>m'$

i.e, $kg(x) \leq (-\frac{N}{k}).k$ $=(-N)$ whenever $x>m'$

Answer 2

Let $k>0$. The equality $$\lim_{x\to \infty} g(x)=-\infty$$ means that for every $M<0$ there exists some $R>0$ such that $$g(x)<\frac{M}{k}<0$$ whenever $x>R$. Now, if we multiply the last inequality by $k$ we get $$kg(x)<M$$ which means, the number $kg(x)$ stays under $M<0$ for any choice of $ M $ as long as $ x> R $. Which means also : $$\lim_{x\to \infty} kg(x)=-\infty$$