Epsilon delta proof for divergence as $ x \rightarrow \infty $

Question

Im having a problem that looks as follows:

Show that $x^5 - x^2 + 3 \rightarrow \infty$ as $x \rightarrow \infty$

My question is how do we construct the proof for such statements? I cannot even find the correct definition for such limits. I know the definition when we have $\lim f(x) = \infty $, but here we also have x approaching $\infty$. All help would be appreciated.

Answer 1

We say that $\lim_{x\to\infty}f(x)=\infty$ when$$(\forall M\in\mathbb R)(\exists N\in\mathbb R):x>N\implies f(x)>M.$$So, take $M\in\mathbb R$. Then, take $N\in\mathbb R$ such that $x>N\implies x^2>M$ and $x^3>2$. Then$$x^5-x^2+3=x^2(x^3-1)+3>M\times1+3>M.$$

Answer 2

The definition of $\lim_{x\to\infty}f(x)=\infty$ is

For any real number $F$, there is a real number $X$ such that $x>X$ implies $f(x)>F$.

In this case we could, for instance, say

If $F<0$, set $X=0$. Otherwise, set $X=F$.

and then prove that this works. Indeed, since our chosen $X$ is non-negative, which means that $x$ will be positive, we have $x^5-x^2+3>x> F$:

• For $x\in [0,1)$, $1>x^2$ gives $x^5-x^2+3\geq 2>x$
• For $x\in [1,2]$, $x^5\geq x^2$ gives $x^5-x^2+3\geq3>x$
• For $x>2$, $x^5=x^3\cdot x^2> 8x^2$ gives $x^5-x^2+3> 7x^2+3>x$

Answer 3

Option:

Let $x>1$;

$f(x):= x^2(x^3-1) +3>$

$ x^2 \cdot 1 +3> x^2 >x$;

Show that

$\lim_{x \rightarrow\infty}f(x)=\infty$, i.e.

for $M$, real, positive $(>1)$, there exists a $N$, real positive, s.t.

$x >N$ implies $f(x)> M$.

Choose $N \ge M$ ( $N >1$), then

$x >N$ implies $f(x)>x > N \ge M$.