Honors Calculus I - 2.2 Limits of Functions

Function Limit

Definition of $\lim_{x\to a}f(x)=L$

$$\forall \varepsilon >0, \;\exist \delta >0,\;\forall x:\,0<|x-a|<\delta\\ |f(x)-L|<\varepsilon$$

The limit of function $f$ at $a$ equals $L$.

Cauchy criterion:

For every $\varepsilon >0$, there exists a $\delta >0$ such that $|f(x)-f(y)|<\varepsilon$ for all $x,y$ satisfying $0<|x-a|<\delta, 0<|x-y|<\delta$

Heine Theorem

The limit of the function $f(x)$ as $x$ approached $a$ is $L$ iff for every sequence $\{x_n\}$, with $x_n\neq a$ for sufficiently large $n$ that converges to $a$, the sequence of function values $\{f(x_n)\}$ converges to $L$.

Rules

• Arithmetic Rules
• The order Rules
• Squeeze Theorem

The composition Rule

Suppose $\lim_{ y\to b}f(y)=L,\lim_{x\to a}g(x)=b$, then

$$\lim_{x\to a}f(g(x))=L$$

provided at least one of the following holds:

1. $f(b)=L$
2. $g(x)\neq b$ near $x=a$ except possibly at $a$

Interpretation:

The order rule for function limits

Suppose that $f$ and $g$ with :

$$\lim_{x\to a}f(x)=A\;\;\lim_{x\to a}g(x)=B$$

1. If $A<B$, then $\exist \delta$ that $\forall |x-a|<\delta$, $f(x)<g(x)$
2. If there is a $\delta_0$ that $\forall |x-a|<\delta_0$ that satisfy $f(x)\le g(x)$, then $A\le B$

Remark: If $f(x)<g(x)$ near $a$, it is incorrect to conclude that $A<B$.

Squeeze Theorem for function limits

If exists $\delta_0$ that $\forall x:|x-a|<\delta$ :

$$g(x)\le f(x)\le h(x)$$

Then if $\lim_{x\to a}g(x)=\lim_{x\to a}h(x)=L$, then $\lim_{x\to a}f(x)=L$.

One Sided Limits

• Left-hand limit : For $\forall \varepsilon >0$, there $\exist \delta > 0$ that $\forall x:a -\delta<x<a$, then $|f(x)-L|<\varepsilon$, we can write in $\lim_{x\to a^{-}}f(x)=L$
• Right-hand limit : same but $a<x<a+\delta$. Write in $\lim_{x\to a^+}f(x)=L$

The limit $\lim_{x\to a}f(x)=L$ iff both one-sided limits $\lim_{x\to a^-}f(x)=\lim_{x\to a^+}f(x)=L$

Some basic limits of functions

• $\lim_{x\to a}x^p=a^p$
• $\lim_{x\to a}b^x=b^a$, where $b>0$
• $\lim_{x\to a}\log_b x=\log_b a$, where $0<b\neq 1$
• $\lim_{x\to a}\sin x=\sin a$, $\lim_{x\to a}\cos x=\cos a$
• $\lim_{x\to a}\dfrac{\sin x}x=1$
• $\lim_{x\to 0}(1+x)^{\frac 1 x}=e$

Limits in the Extended Real Number System

Neighborhood of infinite:

For any real $c$, the set of real numbers $x$ such that $x>c$ is called a neighborhood of $\infty$ and is written $(c,\infty)$. Similarly, the set $(-\infty,c)$ is a neighborhood of $-\infty$.

Suppose, $p,q\in \overline {\R}$, we denote $\lim_{x\to p}f(x)=q$ iff:

for every neighborhood $U$ of $q$, there is a neighborhood $V$ of $p$ such that $f(x)\in U$ for all $x\in V,x\neq p$

• Uniqueness

• Cauchy Criterion also applies

• In fact, the neighborhoods of $\infty$ and $-\infty$ are inherently one sided

• The order rule, arithmetic rules, the composition rules, and the Squeeze Theorem also holds.

Limits at infinity

When $p$ is infinite and $q$ is finite:

$\lim_{x\to \infty}f(x)=q$ or $\lim_{x\to -\infty}f(x)=q$

• horizontal asymptote
  • $\lim_{x\to \infty}f(x)=b$ or $\lim_{x\to -\infty}f(x)=b$
• slant asymptote
  • $\lim_{x\to \infty}[f(x)-(mx+b)]=0$ or $\lim_{x\to -\infty}[f(x)-(mx+b)]=0$

Infinite Limits

If $p$ is finite and $q$ is infinite:

$$\lim_{x\to p}f(x)=\infty$$

Vertical asymptote:

It must be noted that a zero of the denominator $\nLeftrightarrow$a vertical asymptote.