Lecture 20:

Limit Theorems

Proposition. \(\lim\left(\dfrac{n}{n^2+1}\right)=0.\)

Proof. Let \(\varepsilon>0\) be given. By the Archimedean Property there is a number \(K\in \N\) such that \(K>\frac{1}{\varepsilon}+1\). Set \(K(\varepsilon) = K\). Let \(n\geq K(\varepsilon)\) be arbitrary. This implies \(n>\frac{1}{\varepsilon}\), and hence
\[\frac{1}{n}<\varepsilon.\]
For our \(n\) we have
\[\left|\frac{n}{n^2+1} - 0\right| = \frac{n}{n^2+1}\leq \frac{n}{n^2} = \frac{1}{n} <\varepsilon.\]
Therefore, for each \(\varepsilon>0\) we have shown that there exists a natural number \(K(\varepsilon)\) such that
\[\left|\frac{n}{n^2+1} - 0\right|<\varepsilon\text{ for all }n\geq K(\varepsilon).\]
This completes the proof. \(\Box\)

If you are trying to SHOW that \(\lim(x_{n})=x\):

 

For any \(\varepsilon>0\) you must PRODUCE a natural number \(K(\varepsilon)\) such that \[|x_{n}-x|<\varepsilon \text{ for all } n\geq K(\varepsilon).\]

If you are TOLD that \(\lim(x_{n}) = x\):

 

Then given any positive number \(\varepsilon\) you GET a natural number \(K(\varepsilon)\) such that 

\[|x_{n}-x|<\varepsilon \text{ for all } n\geq K(\varepsilon).\]

Proving vs. Using \(\lim(x_{n}) = x\)

Theorem 1. If \(\lim(x_{n}) = x\) and \(\lim(x_{n}) = y\), then \(x=y\).

Theorem 1. If \(\lim(x_{n}) = x\) and \(\lim(x_{n}) = y\), then \(x=y\).

Proof.  Assume toward a contradiction that \(x\neq y\). 

Theorem 1. If \(\lim(x_{n}) = x\) and \(\lim(x_{n}) = y\), then \(x=y\).

Proof.  Assume toward a contradiction that \(x\neq y\). This means that \(\frac{1}{4}|x-y|>0\). 

Theorem 1. If \(\lim(x_{n}) = x\) and \(\lim(x_{n}) = y\), then \(x=y\).

Proof.  Assume toward a contradiction that \(x\neq y\). This means that \(\frac{1}{4}|x-y|>0\). From the assumption that \(\lim(x_{n})=x\) we observe that there is a number \(K\in\N\) such that

\[|x_{n}-x|<\frac{1}{4}|x-y|\text{ for all }n\geq K.\]

 

Theorem 1. If \(\lim(x_{n}) = x\) and \(\lim(x_{n}) = y\), then \(x=y\).

Proof.  Assume toward a contradiction that \(x\neq y\). This means that \(\frac{1}{4}|x-y|>0\). From the assumption that \(\lim(x_{n})=x\) we observe that there is a number \(K\in\N\) such that

\[|x_{n}-x|<\frac{1}{4}|x-y|\text{ for all }n\geq K.\]

From the assumption that \(\lim(x_{n})=y\) we observe that there is a number \(M\in\N\) such that

\[|x_{n}-y|<\frac{1}{4}|x-y|\text{ for all }n\geq M.\]

 

Theorem 1. If \(\lim(x_{n}) = x\) and \(\lim(x_{n}) = y\), then \(x=y\).

Proof.  Assume toward a contradiction that \(x\neq y\). This means that \(\frac{1}{4}|x-y|>0\). From the assumption that \(\lim(x_{n})=x\) we observe that there is a number \(K\in\N\) such that

\[|x_{n}-x|<\frac{1}{4}|x-y|\text{ for all }n\geq K.\]

From the assumption that \(\lim(x_{n})=y\) we observe that there is a number \(M\in\N\) such that

\[|x_{n}-y|<\frac{1}{4}|x-y|\text{ for all }n\geq M.\]

Take some \(n_{0}\in\N\) such that \(n_{0}\geq K\) and \(n_{0}\geq M.\) Then we have

\[|x_{n_{0}}-x|<\frac{1}{4}|x-y|\quad\text{and}\quad |x_{n_{0}}-y|<\frac{1}{4}|x-y|.\]

 

Theorem 1. If \(\lim(x_{n}) = x\) and \(\lim(x_{n}) = y\), then \(x=y\).

Proof.  Assume toward a contradiction that \(x\neq y\). This means that \(\frac{1}{4}|x-y|>0\). From the assumption that \(\lim(x_{n})=x\) we observe that there is a number \(K\in\N\) such that

\[|x_{n}-x|<\frac{1}{4}|x-y|\text{ for all }n\geq K.\]

From the assumption that \(\lim(x_{n})=y\) we observe that there is a number \(M\in\N\) such that

\[|x_{n}-y|<\frac{1}{4}|x-y|\text{ for all }n\geq M.\]

Take some \(n_{0}\in\N\) such that \(n_{0}\geq K\) and \(n_{0}\geq M.\) Then we have

\[|x_{n_{0}}-x|<\frac{1}{4}|x-y|\quad\text{and}\quad |x_{n_{0}}-y|<\frac{1}{4}|x-y|.\]

Finally, by the triangle inequality we have

\[|x-y| = |(x_{n_{0}}-y) - (x_{n_{0}}-x)|\leq |x_{n_{0}}-y| + |x_{n_{0}}-x|\]

\[<\frac{1}{4}|x-y| + \frac{1}{4}|x-y| = \frac{1}{2}|x-y|\]

Theorem 2. If \(\lim(x_{n}) = x\) and \(\lim(x_{n}) = y\), then \(x=y\).

Proof continued. Therefore, 

\[|x-y|< \frac{1}{2}|x-y|.\]

Since \(|x-y|>0\) this implies \(1<\frac{1}{2}\), a contradiction. Hence it must be the case that \(x=y\). \(\Box\)

 

Theorem 2. The sequence \((x_{n})\) has limit \(x\) if and only if the sequence \(\big(|x_{n} - x|\big)\) has limit \(0\). 

Proof. Suppose \((x_{n})\) has limit \(x\). Let \(\varepsilon>0\) be given. By the definition of a limit there is a natural number \(K(\varepsilon)\) such that

\[|x_{n}-x|<\varepsilon\text{ for all }n\geq K(\varepsilon).\]

We will take this to be our \(K(\varepsilon)\) as well. Let \(n\geq K(\varepsilon)\) be arbitrary, then we have

\[\big||x_{n} - x| - 0\big| = |x_{n}-x|<\varepsilon.\]

This shows that \((|x_{n} - x|)\) has limit \(0\).

 

Theorem 2. The sequence \((x_{n})\) has limit \(x\) if and only if the sequence \(\big(|x_{n} - x|\big)\) has limit \(0\). 

Proof. Suppose \((x_{n})\) has limit \(x\). Let \(\varepsilon>0\) be given. By the definition of a limit there is a natural number \(K(\varepsilon)\) such that

\[|x_{n}-x|<\varepsilon\text{ for all }n\geq K(\varepsilon).\]

We will take this to be our \(K(\varepsilon)\) as well. Let \(n\geq K(\varepsilon)\) be arbitrary, then we have

\[\big||x_{n} - x| - 0\big| = |x_{n}-x|<\varepsilon.\]

This shows that \((|x_{n} - x|)\) has limit \(0\).

Suppose \((|x_{n} - x|)\) has limit \(0\). Let \(\varepsilon>0\) be given. By the definition of a limit there is a natural number \(K(\varepsilon)\) such that

\[\big||x_{n} - x| - 0\big|<\varepsilon\text{ for all }n\geq K(\varepsilon).\]

Hence, if \(n\geq K(\varepsilon)\) then

\[|x_{n}-x|= \big||x_{n} - x| - 0\big|<\varepsilon.\]

This shows that \((x_{n})\) has limit \(x\). \(\Box\)

Theorem 3. Assume \(x_{n}\) and \(y_{n}\) are sequences. If

\[0\leq x_{n}\leq y_{n}\text{ for all }n\in\N\]

and \(\lim(y_{n}) = 0\), then \(\lim(x_{n}) = 0.\)

Proof. Let \(\varepsilon>0\) be given. 

Theorem 3. Assume \(x_{n}\) and \(y_{n}\) are sequences. If

\[0\leq x_{n}\leq y_{n}\text{ for all }n\in\N\]

and \(\lim(y_{n}) = 0\), then \(\lim(x_{n}) = 0.\)

Proof. Let \(\varepsilon>0\) be given. By the definition of a limit there is a number \(K(\varepsilon)\in\N\) such that 

\[|y_{n}|<\varepsilon\text{ for all }n\geq K(\varepsilon).\]

 

Theorem 3. Assume \(x_{n}\) and \(y_{n}\) are sequences. If

\[0\leq x_{n}\leq y_{n}\text{ for all }n\in\N\]

and \(\lim(y_{n}) = 0\), then \(\lim(x_{n}) = 0.\)

Proof. Let \(\varepsilon>0\) be given. By the definition of a limit there is a number \(K(\varepsilon)\in\N\) such that 

\[|y_{n}|<\varepsilon\text{ for all }n\geq K(\varepsilon).\]

Since \(y_{n}\geq 0\) we see that

\[y_{n}<\varepsilon\text{ for all }n\geq K(\varepsilon).\]

Let \(n\geq K(\varepsilon)\) be arbitrary. Since \(x_{n}\geq 0\) we have

\[|x_{n} - 0| = x_{n}\leq y_{n}<\varepsilon.\]

Therefore \(\lim(x_{n}) = 0\). \(\Box\)

Example. We have already seen (Example 3.1.6 (a) in the text) that

\[\lim\left(\frac{1}{n}\right) = 0.\]

Consider the sequence

\[\left(\frac{n}{n^{2}+1}\right).\]

Note that for any \(n\in\N\) we have

\[0\leq \frac{n}{n^2+1}\leq \frac{n}{n^2} = \frac{1}{n}.\]

So, by the previous Theorem 

\[\lim\left(\frac{n}{n^{2}+1}\right) = 0.\]

Practice Problem. Prove the following proposition

Proposition. If \(\lim(x_{n}) = x\), then \(\lim(x_{n}+3) = x+3\).