Prove, using the definition of $\lim\limits_{x \to a} f(x) = \infty $, that
$\lim\limits_{x \to 0} \frac{4x+\sqrt{5}}{2x^3+x^2} = \infty$
We know the definiton as:
$\lim\limits_{x \to a} f(x) = \infty $
if, for all $K > 0$, there exists $\delta > 0$, such that $\lvert f(x)\rvert > K$, when $0 < \lvert x -a\rvert < \delta$
In our case:
$f(x) = \frac{4x+\sqrt{5}}{2x^3+x^2}$
$a = 0$
From the defintion:
$\frac{4x+\sqrt{5}}{2x^3+x^2} > K$, when $\lvert x -a\rvert = \lvert x \rvert <\delta$
Where do we go from here?
Observe that $4x+\sqrt{5}>4x$ so
$$\frac{4x+\sqrt{5}}{2x^3+x^2} > \frac{4x}{2x^3+x^2}>K$$ if $$\frac{4}{2x^2+x}>K$$
we know $K>0;\;x>0$ so the inequality is solved when $$0<x<\frac{1}{4} \sqrt{\frac{K+32}{K}}-\frac{1}{4}$$
So for any $K>0$ if $$|x|<\delta=\frac{1}{4} \sqrt{\frac{K+32}{K}}-\frac{1}{4}$$ then $f(x)>K$
For instance if $K=1000$ then $\delta=0.0039$
Taking $x=0.001<\delta$ we have $$f(x)=\frac{4\cdot 0.001+\sqrt{5}}{2\cdot 0.001^3+0.001^2}\approx 2.23\times 10^6>K$$
Hope this helps