Using the definition of $\lim_{x\to a} f(x) = \infty$ , prove that $\lim_{x\to 0} \frac{4x+\sqrt{5}}{2x^3+x^2} = \infty$

Question

This is my working so far:

$$\lim_{x\to a} f(x) = \infty$$ if,

$\forall K > 0, \, \exists \delta >0$ s.t. $|f(x)|>K$ when $0<|x-a|<\delta$ & $x\in \text{Dom}(f)$

Proof: Given $K>0$, find $\delta>0$ s.t. $|f(x)| > K$ when $0<|x-a|<\delta$

i.e. that, $$\left|\frac{4x+\sqrt{5}}{2x^3+x^2}\right| > K\quad \text{ when }|x|<\delta$$

My problem begins here in that I haven's a clue how to simplify $\left|\frac{4x+\sqrt{5}}{2x^3+x^2}\right| > K$ and get something in terms of $x$ or $|x|$ to then determine my $\delta$ accordingly.

Thank you for taking the time and $\textbf{hints only please!}$

Answer 1

Hint

We have that

$$\left|\frac{4x+\sqrt{5}}{2x^3+x^2}\right|=\dfrac{1}{x^2}\left|\frac{4x+\sqrt{5}}{2x+1}\right|.$$ Assume $|x|< 1/4.$ Then

$$|4x+\sqrt{5}|> \sqrt{5}-1$$ and $$|2x+1|< \dfrac12. $$

Answer 2

If $|x| < 1/4$, numerator > 1 and |denominator| $=|x^2(2x+1)| < |x^2/2|$.

Answer 3

Then, $|x|<\delta\le\frac{1}{4}$ and for $\delta=\sqrt{\dfrac{\sqrt 5-1}{2K}}$ we have that $|f(x)|>\dfrac{\sqrt 5-1}{2x^2}>K$, with the restriction that $\delta\le\dfrac{1}{4}$ leading to the restriction $K>8(\sqrt5-1)$ which is definitely $\textbf{not}$ $\forall K>0$. This solution is $\textbf{wrong}$. Delete this if you consider it necessary, but please mention this in comments as I cannot do it. Setting random boundaries does not work in general. It leads to restrictions that cannot be handled.