Function/R/Limit/Epsilon/Introduction/Section

Quite often, functions are not defined in certain points, e.g. if the used functional term is not defined there. However, it makes a huge difference whether only the functional term is not defined in a point, but has a useful (continuous) extension, or whether the function does not have a useful extension in this point, e.g. because it has a pole or an even more chaotic behavior. The following concept is in particular relevant for the definition of differentiability (if the difference quotient has a useful limit, then it is called differential quotient).

Definition

Let ${}T\subseteq \mathbb {R}$ denote a subset and ${}a\in \mathbb {R}$ a point. Let

$f\colon T\longrightarrow \mathbb {R}$

be a function. Then ${}b\in \mathbb {R}$ is called limit of ${}f$ in ${}a$, if for every ${}\epsilon >0$ there exists some ${}\delta >0$ such that for all ${}x\in T$ fulfilling

${}\vert {x-a}\vert \leq \delta \,,$

the estimate

${}\vert {f(x)-b}\vert \leq \epsilon \,$

holds. In this case, we write

${}\operatorname {lim} _{x\rightarrow a}\,f(x)=b\,.$

This concept is basically only useful if there exists at least some sequence within ${}T$ converging to ${}a$. A typical situation is the following: Let ${}I$ denote a real interval, ${}a\in I$ a point and let ${}T=I\setminus \{a\}$. The function is defined on ${}T$ but not in ${}a$, and we are dealing with the question whether ${}f$ can be extended to a function ${}{\tilde {f}}$ defined on the whole ${}I$. Here, ${}{\tilde {f}}(a)$ should be determined by ${}f$.

Lemma

Let ${}T\subseteq \mathbb {R}$ denote a subset and ${}a\in \mathbb {R}$ a point. Let

$f\colon T\longrightarrow \mathbb {R}$

be a function and ${}b\in \mathbb {R}$

a point. Then the following statements are equivalent.

1. We have

   ${}\operatorname {lim} _{x\rightarrow a}\,f(x)=b\,.$

2. For every sequence ${}{\left(x_{n}\right)}_{n\in \mathbb {N} }$ in ${}T$ which converges to ${}a$, also the image sequence ${}{\left(f(x_{n})\right)}_{n\in \mathbb {N} }$ converges to ${}b$.

Proof

See exercise.

$\Box$

This implies for a continuous function ${}f\colon T\rightarrow \mathbb {R}$ that it can be extended to a continuous function ${}{\tilde {f}}\colon T\cup \{a\}\rightarrow \mathbb {R}$ (by ${}{\tilde {f}}(a)=b$) if and only if the limit of ${}f$ in ${}a$ equals ${}b$.

Lemma

Let ${}T\subseteq \mathbb {R}$ denote a subset and ${}a\in \mathbb {R}$ a point. Let ${}f\colon T\rightarrow \mathbb {R}$ and ${}g\colon T\rightarrow \mathbb {R}$ denote functions, such that the limits ${}\operatorname {lim} _{x\rightarrow a}\,f(x)$ and ${}\operatorname {lim} _{x\rightarrow a}\,g(x)$

exist. Then the following statements hold.

1. The sum ${}f+g$ has in ${}a$ the limit

   ${}\operatorname {lim} _{x\rightarrow a}\,(f(x)+g(x))=\operatorname {lim} _{x\rightarrow a}\,f(x)+\operatorname {lim} _{x\rightarrow a}\,g(x)\,.$

2. The product ${}f\cdot g$ has in ${}a$ the limit

   ${}\operatorname {lim} _{x\rightarrow a}\,(f(x)\cdot g(x))=\operatorname {lim} _{x\rightarrow a}\,f(x)\cdot \operatorname {lim} _{x\rightarrow a}\,g(x)\,.$

3. Suppose that ${}g(x)\neq 0$ for all ${}x\in T$ and ${}\operatorname {lim} _{x\rightarrow a}\,g(x)\neq 0$. Then the quotient ${}f/g$ has in ${}a$ the limit

   ${}\operatorname {lim} _{x\rightarrow a}\,{\frac {f(x)}{g(x)}}={\frac {\operatorname {lim} _{x\rightarrow a}\,f(x)}{\operatorname {lim} _{x\rightarrow a}\,g(x)}}\,.$

Proof

This follows from fact and from fact.

$\Box$

Example

We consider the limit

$\operatorname {lim} _{x\rightarrow 0}\,{\frac {{\sqrt {x+4}}-2}{x}},$

where ${}x\in \mathbb {R} \setminus \{0\},\,x\geq -4$. For ${}x=0$, this term is not defined, and from this term one can not read of directly whether the limit exists. It is however possible to multiply the numerator and the denominator by ${}{\sqrt {x+4}}+2$, then we get

${}{\begin{aligned}{\frac {{\sqrt {x+4}}-2}{x}}&={\frac {{\left({\sqrt {x+4}}-2\right)}{\left({\sqrt {x+4}}+2\right)}}{x{\left({\sqrt {x+4}}+2\right)}}}\\&={\frac {x+4-4}{x{\left({\sqrt {x+4}}+2\right)}}}\\&={\frac {x}{x{\left({\sqrt {x+4}}+2\right)}}}\\&={\frac {1}{{\sqrt {x+4}}+2}}.\end{aligned}}$

Due to the rules for limits, we can determine the limit in the numerator and in the denominator separately, where for the denominator we use the continuity of the square root according to exercise. Hence, the limit is ${}1/4$.