As the title says, I'm trying to find a suitable $\delta$ to complete the proof of $$\lim_{x\to 0} \frac{{\cos}^2(x)-1}{x^4} = -\infty.$$
Starting out with $$\forall N<0, \exists \delta >0 \text{ s.t. } 0<\lvert x-0\rvert \implies \frac{{\cos}^2(x)-1}{x^4}<N,$$ I thought setting $\delta \leq 1$ could be a good idea, leading to the restriction $-1<x<1$. Then I tried some basic algebraic manipulation like $$\frac{{\cos}^2(x)-1}{x^4}=-\frac{{\sin}^2(x)}{x^4} < - \frac{{\sin}^2(x)}{\delta^4},$$ but that lead me nowhere. So any help or hints would be appreciated.
Since $$ \lim_{x\to0}\frac{\sin x}{x}=1 $$ one can choose $\delta_1>0$ such that when $|x|<\delta_1$ $$ \frac{\sin x}{x}\ge \frac12. $$ So when $|x|<\delta_1$, one has $$\frac{{\cos}^2(x)-1}{x^4}=-\frac{{\sin}^2(x)}{x^4} =-\frac{{\sin}^2(x)}{x^2}\frac{1}{x^2}< - \frac{1}{2x^2}. $$ For any given $N>0$, letting $- \frac{1}{2x^2}<-N$ gives $|x|<\frac{1}{\sqrt{2N}}$. Define $\delta=\min\{\delta_1, \frac{1}{\sqrt{2N}}\}$. Thus when $0<|x|<\delta$, $$\frac{{\cos}^2(x)-1}{x^4}<-N. $$ Namely $$\lim_{x\to0}\frac{{\cos}^2(x)-1}{x^4}=-\infty. $$