Prove using the formal definition of a limit that $\lim_{x\to 0}\frac{3x+\sqrt 7}{x^3 + x^2}=\infty$

Question

I know that given $K>0$ we must search for $\delta>0$ such that $|f(x)|>k$ when $|x|<\delta$ but cannot come up with an entire proof.

Answer 1

Because $x\to 0$ we can assume $|x| < \frac{1}{3}$ hence $$\frac{3x+\sqrt 7}{x^3 + x^2} \ge \frac{\sqrt{7} - 1}{x^3 + x^2} \ge \frac{1}{2x^2}$$

The rest you can do by your own?