Formal limit definition in $\lim\limits_{x\to+\infty} \frac{x^2+2}{9x}=+\infty$

Question

Using the limit definition, show that:

$\lim\limits_{x\to+\infty} \frac{x^2+2}{9x}=+\infty$

I get blocked when I use the equation:

Definition

$\lim\limits_{x\to+\infty} f(x)=+\infty$

$f(x)$ tends to $+\infty$ as $x$ tends to $ \infty$ if and only if $$ \forall A>0, \exists B>0 \ \mbox{ such that} \ \forall x \ \mbox{where} \ x>B, \\ f(x)>A$$

but I do not know how to continue

Answer 1

For $x>0$ we have that

$$\dfrac{x^2+2}{9x}>\dfrac{x^2}{9x}=\dfrac{x}{9}.$$

Thus

$$x>B\implies \dfrac{x^2+2}{9x}>\dfrac{x}{9}=\dfrac{B}{9}.$$

So, consider $B=9A$ and you are done.

Answer 2

For a given $A>0$, let $B = 9A$. Then we have that:

$$x > 9A \implies f(x) = \frac{x^2+2}{9x} = \frac x9 + \frac{2}{9x} > \frac x9 > A$$

Hence the proof.