Limits at Infinity

See Limits of Functions.

Definition

Let $f:\mathbb{R}\to \mathbb{R}$.

1. ${\lim }_{x\to \infty }f(x)=\ell$ if $\exists \ell \in R$ such that $\exists \ell \in \mathbb{R},\forall \epsilon >0,\exists m\in \mathbb{R}$ such that $x>M$, then $|f(x)-\ell |<\epsilon$.
2. Let $p\in \mathbb{R}$. We say ${\lim }_{x\to p}f(x)=\infty$ if $\forall M\in \mathbb{R},\exists \delta >0$ such that $0<|x-p|<\delta ⟹f(x)>M$.
3. ${\lim }_{x\to \infty }f(x)=\infty$ if $\forall M\in \mathbb{R},\exists N\in \mathbb{R}$ such that $\forall x>N$, then $f(x)>M$. These are all similar for $-\infty$.

Example 1

$$\underset{x\to 0}{\lim }(1/x^2)=+\infty$$

Proof: Let $M\in \mathbb{R}$. Need to find $\delta >0$ such that $0<|x-0|<\delta$ we have $1/x^2>0$.

1. $0<|x|<\delta ⟹1/x^2>1/{\delta }^2$
2. If $M\le 0$, for $\delta =1$, then $f(x)=1/x^2\ge 0\ge M$.
3. If $M>0$, let $M=1/{\delta }^2⟺\delta =\sqrt{M}$.

So if $0<|x|<\delta$, then $f(x)=1/x^2>1/{\delta }^2=M$.