Real Function Limits:Sequential Criterion

We will now look at a very important theorem known as The Sequential Criterion for a Limit which merges the concept of the limit of a function $f$ at a cluster point $c$ from $A$ with regards to sequences $(a_{n})$ from $A$ that converge to $c$.

Theorem 1 (The Sequential Criterion for a Limit of a Function): Let $f:A\to \mathbb {R}$ be a function and let $c$ be a cluster point of $A$. Then $\lim _{x\to c}f(x)=L$ if and only if for all sequences $(a_{n})$ from the domain $A$ where $a_{n}\neq c$ $\forall n\in \mathbb {N}$ and $\lim _{n\to \infty }a_{n}=c$ then $\lim _{n\to \infty }f(a_{n})=L$.

Consider a function $f$ that has a limit $L$ when $x$ is close to $c$. Now consider all sequences $(a_{n})$ from the domain $A$ where these sequences converge to $c$, that is $\lim _{n\to \infty }a_{n}=c$. The Sequential Criterion for a Limit of a Function says that then that as $n$ goes to infinity, the function $f$ evaluated at these $a_{n}$ will have its limit go to $L$.

For example, consider the function $f:\mathbb {R} \to \mathbb {R}$ defined by the equation $f(x)=x$, and suppose we wanted to compute $\lim _{x\to 0}x$. We should already know that this limit is zero, that is $\lim _{x\to 0}x=0$. Now consider the sequence $(a_{n})=\left({\frac {1}{n}}\right)$. This sequence $(a_{n})$ is clearly contained in the domain of $f$. Furthermore, this sequence converges to 0, that is $\lim _{n\to \infty }{\frac {1}{n}}=0$. If all such sequences $(a_{n})$ that converge to $0$ have the property that $(f(a_{n}))$ converges to $f(0)=0$, then we can say that $\lim _{n\to 0}f(x)=0$.

We will now look at the proof of The Sequential Criterion for a Limit of a Function.

• Proof: $\Rightarrow$ Suppose that $\lim _{x\to c}f(x)=L$, and let $(a_{n})$ be a sequence in $A$ such that $a_{n}\neq c$ $\forall n\in \mathbb {N}$ such that $\lim _{n\to \infty }a_{n}=c$. We thus want to show that $\lim _{n\to \infty }f(a_{n})=L$.

• Let $\varepsilon >0$. We are given that $\lim _{x\to c}f(x)=L$ and so for $\varepsilon >0$ there exists a $\delta >0$ such that if $x\in A$ and $0<\mid x-c\mid <\delta$ then we have that $\mid f(x)-L\mid <\varepsilon$. Now since $\delta >0$, since we have that $\lim _{n\to \infty }a_{n}=c$ then there exists an $N\in \mathbb {N}$ such that if $n\geq N$ then $\mid a_{n}-c\mid <\delta$. Therefore $a_{n}\in V_{\delta }(c)\cap A$.

• Therefore it must be that $\mid f(a_{n})-L\mid <\varepsilon$, in other words, $\forall n\geq N$ we have that $\mid f(a_{n})-L\mid <\varepsilon$ and so $\lim _{n\to \infty }f(a_{n})=L$.

• $\Leftarrow$ Suppose that for all $(a_{n})$ in $A$ such that $a_{n}\neq c$ $\forall n\in \mathbb {N}$ and $\lim _{n\to \infty }a_{n}=c$, we have that $\lim _{n\to \infty }f(a_{n})=L$. We want to show that $\lim _{x\to c}f(x)=L$.

• Suppose not, in other words, suppose that $\exists \varepsilon _{0}>0$ such that $\forall \delta >0$ then $\exists x_{\delta }\in A\cap V_{\delta }(c)\setminus \{c\}$ such that $\mid f(x_{\delta })-L\mid \geq \varepsilon _{0}$. Let $\delta _{n}={\frac {1}{n}}$. Then there exists $x_{\delta _{n}}=a_{n}\in A\cap V_{\delta _{n}}(c)\setminus \{c\}$, in other words, $0<\mid a_{n}-c\mid <{\frac {1}{n}}$ and $\lim _{n\to \infty }a_{n}=c$. However, $\mid f(a_{n})-L\mid \geq \varepsilon _{0}$ so $\lim _{n\to \infty }f(a_{n})\neq L$, a contradiction. Therefore $\lim _{x\to c}f(x)=L$.

Licensing

Content obtained and/or adapted from:

• The Sequential Criterion for a Limit of a Function, mathonline.wikidot.com under a CC BY-SA license