Notes for Limits involving $\infty$

Exercises and solutions are included with Week #3 suggested problems.

Limits TOWARD $\infty$

Recall that the regular limit of $f(x)$ at $c$ is $L$ if (roughly speaking) ``$f(x)$ is near the line $y=L$ when we restrict attention to $x\approx c$ . ''

Idea 1   How can we restrict attention to $x\approx\infty$ ?

• choose a threshold $0<N$
• insist that $N < x$

Definition 1 (Limit as $x\to+\infty$ )   Let function $f(x)$ be defined on at least a ray $(p,+\infty)$ .

We write `` $\lim_{x\to+\infty}f(x) = L$ ''if:

$$\begin{aligned} &&\text{\textquotedblleft Given }&\epsilon > 0,... ... \right\rvert < \epsilon. \text{\textquotedblright} \end{aligned}$$

To avoid $\epsilon - N$ proofs, we have the following,

Theorem 1 (Equivalence for $x\to+\infty$ )  

$$\lim_{x\to+\infty}f(x) = L \Longleftrightarrow \lim_{h\to 0^+}f\left(\frac{1}{h}\right) = L.$$

Remark 1 (Limit as $x\to -\infty$ )   There are the expected analogues:

• [Defn] Let function $f(x)$ be defined on at least a ray $(-\infty,-p)$ .

  We write `` $\lim_{x\to-\infty}f(x) = L$ ''if given $\epsilon > 0$ there is some $N(\epsilon) > 0$ so that,

  $$x < -N(\epsilon) \Longrightarrow \left\lvert f(x)-L \right\rvert < \epsilon.$$

• [Thm] $\lim_{x\to-\infty}f(x) = L \Leftrightarrow \lim_{h\to 0^-}f\left(\frac{1}{h}\right) = L$

Example 1   Prove $\lim_{x\to\infty}\left(\frac{1}{x}+1\right) = 1$ .

Example 2   Find $\lim_{x\to-\infty}\frac{\cos^2(x)+1}{x}$ .

Example 3   Find $\lim_{x\to\infty}e^\frac{1}{x\cos\left(\frac{1}{x}\right)}$ .

Example 4   Find $\lim_{x\to-\infty}\frac{\sqrt{x+x^2}}{3x}$ .

Example 5   Let $f(x)$ be the Dirichlet function below. Prove $\lim_{x\to\infty}f(x)$ does not exist.

$$f(x) = \left\{\begin{aligned} 1 &&& \text{ if } x \text{ rational}\\ 0 &&& \text{ if } x \text{ irrational}\end{aligned}\right.$$

Limits DIVERGING to $\infty$

Recall that the regular limit of $f(x)$ at $c$ is $L$ if (roughly speaking) ``$f(x)$ is near the line $y=L$ when we restrict attention to $x\approx c$ . ''

Idea 2   How can $f(x)$ be ``near $\infty$ ''?

Definition 2 (Limit diverges to $+\infty$ )   Let function $f(x)$ be defined on at least open interval $(c-p,c+p)$ , except perhaps at $c$ .

We write `` $\lim_{x\to c}f(x) = +\infty$ ''if:

$$\begin{aligned} &&\text{\textquotedblleft Given }&M > 0, \text{... ...\Longrightarrow M < f(x). \text{\textquotedblright} \end{aligned}$$

Remark 2 (Don't Exist)   `` $\lim_{x\to c}f(x) = +\infty$ ''is a special way to say limit doesn't exist. Note that, `` $\lim_{x\to c}f(x) \neq \infty$ ''does not mean the limit exists.

To avoid $M-\delta$ proofs, we have the following,

Theorem 2 (Equivalence for $\lim\to\infty$ )  

$$\lim_{x\to c}f(x) = \infty \Longleftrightarrow \begin{aligned... ...e interval } (c-p,c+p) \text{ except perhaps at }c. \end{aligned}$$

Remark 3 (Limit diverges to $-\infty$ )   There are the expected analogues:

• [Defn] Let function $f(x)$ be defined on at least $(c-p,c+p)$ except perhaps at $c$ .

  We write `` $\lim_{x\to c}f(x) = -\infty$ ''if given $M > 0$ there is some $\delta(M) > 0$ so that,

  $$0<\left\lvert x-c \right\rvert <\delta \Longrightarrow f(x) < - M.$$

• [Thm] $\lim_{x\to c}f(x) = -\infty \Leftrightarrow \lim_{x\to c}\frac{1}{f(x)} = 0$ and $f(x) < 0$ on some interval $(c-p,c+p)$ .

There are also the four expected one-sided analogues.

There are also the four combinations of ``diverging to $\infty$ as approach $\infty$ ''.

Example 6   Determine $\lim_{x\to 1}\frac{x^3-1}{\left(x-1\right)^2}$ .

Example 7   Determine $\lim_{x\to 0}\frac{x^2+3x-1}{2-\sqrt{x^2+4}}$ .