Lecture 17:
More limits of sequences
Definition. A sequence \(X = (x_{n})\) in \(\R\) is said to converge to \(x\in\R\), or \(x\) is said to be the limit of \((x_{n})\), if for every \(\varepsilon>0\) there exists a natural number \(K(\varepsilon)\) such that for all \(n\geq K(\varepsilon)\), the terms \(x_{n}\) satisfy
\(|x_{n}-x|<\varepsilon\).
Let's look at that definition again
We want to find a natural number \(K(\varepsilon)\) so that this quantity is small for \(n\geq K(\varepsilon)\)
Prove that \(\displaystyle{\lim\left(\frac{n^2}{n^2+2n+1}\right)=1}\).
Scratch work. First, we look at the quantity that we want to be small
\[\left|\frac{n^2}{n^2+2n+1} - 1\right| = \left|\frac{n^2-(n^2+2n+1)}{n^2+2n+1}\right| = \left|\frac{-2n-1}{n^2+2n+1}\right|\]
Since \(n\in\N\) we see that \(n^2+2n+1>0\). We also see that \(-2n-1\leq -3\). We conclude that \(\frac{-2n-1}{n^2+2n+1}<0\) and hence
\[\left|\frac{-2n-1}{n^2+2n+1}\right| = \frac{2n+1}{n^2+2n+1} <\frac{2n+1}{n^2} \leq \frac{3n}{n^2} = \frac{3}{n}.\]
We want \(\frac{3}{n}<\varepsilon\), and thus we need to take \(n>\frac{3}{\varepsilon}\).
Thus, it is enough to take \(K(\varepsilon)>\frac{3}{\varepsilon}.\)
Proposition. If \(a>0\) and \(c>d>0\), then
\[\frac{a}{c}<\frac{a}{d}.\]
This statement says, provided the numerator and denominators are all positive, if two fractions have the same numerator, then the one with the bigger denominator is the smaller fraction.
Be careful: If \(n\in\N\) then it is true that \(n>n-1\), but it is not true that \(\dfrac{1}{n}<\dfrac{1}{n-1}\), since \(n-1\) is not necessarily positive!
Always be certain that all numerators and denominators are positive.
Proof. Suppose \(c>d>0\). This implies that both \(\frac{1}{c}\) and \(\frac{1}{d}\) are positive, and hence their product is positive. Multiplying both sides of \(c>d\) by the positive number \(\frac{1}{cd}\) we obtain \(\frac{1}{d}>\frac{1}{c}\). Since \(a>0\), multiplying both sides of this inequality by \(a\) we have
\[\frac{a}{d}>\frac{a}{c}.\quad\Box\]
Here's the work I scribble out when I want to remember why this is true:
- \(c>d>0\ \Rightarrow \frac{1}{c}>0\text{ and }\frac{1}{d}>0\)
- Mult. \(c>d\) by both \(\frac{1}{c}\) and \(\frac{1}{d}\) and we get \(\frac{1}{d}>\frac{1}{c}\).
- Mult. by \(a\) and get \(\frac{a}{d}>\frac{a}{c}\)
Proof. Let \(\varepsilon>0\) be given. By the Archimedean Property there is a natural number \(K(\varepsilon)> \frac{3}{\varepsilon}\). Let \(n\in\N\) such that \(n\geq K(\varepsilon)\) be given. By the choice of \(K(\varepsilon)\) we have \(n>\frac{3}{\varepsilon}\) and therefore
\[\frac{3}{n}<\varepsilon.\]
Since \(n\in\N\) we have \(n\geq 1\) and hence \(n^2+2n+1>0 \). We also have \(-2n-1\leq -3\) and hence
\[\left|\frac{-2n-1}{n^2+2n+1}\right| = \frac{2n+1}{n^2+2n+1} <\frac{2n+1}{n^2} \leq \frac{3n}{n^2} = \frac{3}{n}<\varepsilon\]
Finally, we can deduce the following
\[\left|\frac{n^2}{n^2+2n+1} - 1\right| = \left|\frac{n^2-(n^2+2n+1)}{n^2+2n+1}\right| = \left|\frac{-2n-1}{n^2+2n+1}\right|<\varepsilon.\ \Box\]
Prove that \(\displaystyle{\lim\left(\frac{n^2}{n^2+2n+1}\right)=1}\).
Prove that \(\displaystyle{\lim\left(\frac{-n}{n^2-10n+1}\right)=0}\).
Scratch work. First, we look at the quantity that we want to be small
\[\left|\frac{-n}{n^2-10n+1}\right| = \frac{|-n|}{|n^2-10n+1|} = \frac{n}{|n^2-10n+1|}\]
From the following table we see that \(n^2-10n+1\) does not have the same sign for all \(n\in\N\)\[\begin{array}{c|c} n & n^2-10n+1\\ \hline 1 & -8\\ 2 & -15\\ 3 & -20\\ 4 & -23\\ 5 & -24\\ 6 & -23\\ 7 & -20\\ 8 & -15\\ 9 & -8\\ 10 & 1\end{array}\]
Scratch work (continued). If \(n\geq 10\), then \(n^2\geq 10n\), so we have
\[n^2-10n+1\geq 1.\]
Hence, if we make sure \(K(\varepsilon)\geq 10\), and hence \(n\geq 10\), then we have
\[\left|\frac{-n}{n^2-10n+1}\right| = \frac{n}{n^2-10n+1}\leq \frac{n}{n^2-10n} = \frac{1}{n-10}.\]
We need
\[\frac{1}{n-10}<\varepsilon\quad\Leftrightarrow\quad \frac{1}{\varepsilon}<n-10\quad\Leftrightarrow\quad \frac{1}{\varepsilon}+10<n\]
Thus, we will take
\[K(\varepsilon)\geq 11+\frac{1}{\varepsilon}.\]
Just to be extra sure!
Proof. Let \(\varepsilon>0\) be given. By the Archimedean Property there is a natural number \(K(\varepsilon)\geq \frac{1}{\varepsilon}+11\). Let \(n\in\N\) such that \(n\geq K(\varepsilon)\) be given. By the choice of \(K(\varepsilon)\) we have \(n>\frac{1}{\varepsilon}+10\) and therefore
\[\frac{1}{n-10}<\varepsilon\quad\text{and}\quad n>10.\]
Since \(n>10\), it follows that \(n^2>10n\). This implies
\[n^2-10n+1> 1.\]
In particular, \(n^2-10n+1>0\). Finally, we have
\[\left|\frac{-n}{n^2-10n+1}\right| = \frac{n}{n^2-10n+1}\leq \frac{n}{n^2-10n} = \frac{1}{n-10}<\varepsilon.\]
\[ \]
\(\Box\)
Prove that \(\displaystyle{\lim\left(\frac{-n}{n^2-10n+1}\right)=0}\).
Prove that \(\displaystyle{\lim\left(\frac{n^2+1}{n^2-n+3}\right)=1}\).
Scratch work. First, we look at the quantity that we want to be small
\[\left|\frac{n^2+1}{n^2-n+3} - 1\right| = \left|\frac{n^2+1 - (n^2-n+3)}{n^2-n+3} \right| = \left|\frac{n-2}{n^2-n+3} \right|\]
If \(n\geq 3\) then \(n^2\geq 3n\) and \(n^2-n+3\geq 2n+3\geq 9\). We also have \(n-2\geq 1\). Thus \(\frac{n-2}{n^2-n+3}>0\), and
\[ \left|\frac{n-2}{n^2-n+3} \right| = \frac{n-2}{n^2-n+3} < \frac{n}{n^2-n} = \frac{1}{n-1}\]
Hence, it is enough to take \(K(\varepsilon)\) such that both \(K(\varepsilon)>\frac{1}{\varepsilon}+1\), and \(K(\varepsilon)\geq 3\). We can accomplish both by taking \(K(\varepsilon)>3+\frac{1}{\varepsilon}\).
Practice problem: Write out the proof.
Proof. Let \(\varepsilon>0\) be given. By the Archimedean Property there is a natural number \(K(\varepsilon)\geq \frac{1}{\varepsilon}+3\). Let \(n\in\N\) such that \(n\geq K(\varepsilon)\) be given. By the choice of \(K(\varepsilon)\) we have \(n>\frac{1}{\varepsilon}+3>\frac{1}{\varepsilon}+1\) and therefore
\[\frac{1}{n-1}<\varepsilon\quad\text{and}\quad n>3.\]
Since \(n>3\), it follows that \(n^2>3n\). This implies
\[n^2-n+3> 2n+3>9.\]
In particular, \(n^2-n+3>0\). Finally, we have
\[\left|\frac{n^2+1}{n^2-n+3} - 1\right| = \left|\frac{n-2}{n^2-n+3} \right| = \frac{n-2}{n^2-n+3}<\frac{n}{n^2-n} = \frac{1}{n-1}<\varepsilon.\]
\[ \]
\(\Box\)
Prove that \(\displaystyle{\lim\left(\frac{n^2+1}{n^2-n+3}\right)=1}\).
Lecture 17
By John Jasper
Lecture 17
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