Prove using the formal definition of a limit that How would I go about proving this limit?
$\lim_{x\to\infty}\frac{1}{x^4+x^2+5}=0$
so far I have:
$|f(x) -L|< ϵ$ wherever $x > N$
$|\frac{1}{x^4+x^2+5} - 0| < ϵ $ wherever $x > N$
$|\frac{1}{x^4+x^2+5}| < ϵ  ,\; x > ∞$, assuming, $x > 0$
taking the absolute value $\frac{1}{x^4+x^2+5} < ϵ$
${x^4+x^2+5}$ > $\frac{1}{ϵ}$
i am not sure whether this is correct, but what would be the next step?

We have: $x^4+x^2+5 > x^4 \Rightarrow \dfrac{1}{x^4+x^2+5} < \dfrac{1}{x^4}$. Thus you need $\dfrac{1}{x^4} < \epsilon \Rightarrow x^4 > \epsilon^{-1} \Rightarrow x > \epsilon^{-\frac{1}{4}}$. This means for a given $\epsilon > 0$, choose $N$ a natural number such that $N \geq 1+ \epsilon^{-\frac{1}{4}}$, then whenever $x > N \Rightarrow |f(x)| < \epsilon \Rightarrow \displaystyle \lim_{x \to \infty} f(x) = 0$.
