Discrete Mathematics: Prove that f(x) is in O(x)

Question

• Prove that $$\frac{2x^{2}+x}{x+1}$$ is in $O(x)$

Answer 1

By definition, we have that

$$\frac{2x^2+x}{x+1}=\mathcal O(x)\iff \exists\,\text{a constant}\;M\;\;s.t.\;\;\left|\frac{\frac{2x^2+x}{x+1}}x\right|=\left|\frac{2x+1}{x+1}\right|\le M$$

But we know that

$$\lim_{x\to\pm\infty}\frac{2x+1}{x+1}=2\;,\;\;\;\text{ so}\ldots$$

Answer 2

$\frac{2 x^2 +x}{x+1} <\frac{2x^2 +x^2}{x}=3x=O(x)$.