Lecture 16:

Proving \(\lim (x_{n}) = x\)

Proposition 1. \(\displaystyle{\lim\left(\frac{n}{2n+1}\right)=\frac{1}{2}}\).

Scratch work. We wish to show that

\[\left|\frac{n}{2n+1}-\frac{1}{2}\right|<\varepsilon.\]

So, we simplify this inequality as follows:

\[\varepsilon>\left|\frac{n}{2n+1}-\frac{1}{2}\right| = \left|\frac{2n}{4n+2}-\frac{2n+1}{4n+2}\right| = \left|\frac{-1}{4n+2}\right| = \frac{1}{4n+2}\]

Now, we see

\[\frac{1}{4n+2}<\varepsilon\ \Leftrightarrow\ \frac{1}{\varepsilon}<4n+2  \ \Leftrightarrow\ \frac{1}{4}\left(\frac{1}{\varepsilon}-2\right)<n\]

Thus, it seems that we could take \(K(\varepsilon)\) to be a natural number greater than \(\frac{1}{4}\left(\frac{1}{\varepsilon}-2\right)\). Such a natural number exists by the Archimedean Property.

Proposition 1. \(\displaystyle{\lim\left(\frac{n}{2n+1}\right)=\frac{1}{2}}\).

Proof. Let \(\varepsilon>0\) be given. By the Archimedian Property there is a natural number \(K(\varepsilon)\) such that

\[K(\varepsilon)> \frac{1}{4}\left(\frac{1}{\varepsilon}-2\right).\]

Let \(n\in\N\) such that \(n\geq K(\varepsilon)\) be arbitrary. Our choice of \(K(\varepsilon)\) implies \(n> \frac{1}{4}\left(\frac{1}{\varepsilon}-2\right).\) This inequality is equivalent to

\[\frac{1}{4n+2}<\varepsilon.\]

Finally, for this \(n\) we have

\[\left|\frac{n}{2n+1}-\frac{1}{2}\right| = \left|\frac{2n}{4n+2}-\frac{2n+1}{4n+2}\right| = \left|\frac{-1}{4n+2}\right| = \frac{1}{4n+2}<\varepsilon.\ \Box\]

Proposition 1. \(\displaystyle{\lim\left(\frac{3n^2}{n^2+1}\right)=3}\).

Scratch work. First, we simplify the desired inequality:

\[\varepsilon>\left|\frac{3n^2}{n^2+1}-3\right| = \left|\frac{3n^2}{n^2+1}-\frac{3n^2+3}{n^2+1}\right| = \left|\frac{-3}{n^2+1}\right| = \frac{3}{n^2+1}\]

Now, we see

\[\frac{3}{n^2+1}<\varepsilon\ \Leftrightarrow\ \frac{3}{\varepsilon}-1<n^2\]

We want to say that this is equivalent to 

\[\sqrt{\frac{3}{\varepsilon}-1}<n\]

However, if \(\varepsilon>3,\) then \(\frac{3}{\varepsilon}-1<0,\) and so \(\sqrt{\frac{3}{\varepsilon}-1}\) makes no sense. However, \(\frac{3}{\varepsilon}>0\), and thus \(\sqrt\frac{3}{\varepsilon}\) makes sense. Moreover, if \(n>\sqrt{\frac{3}{\varepsilon}}\), then

\[n^2>\frac{3}{\varepsilon}>\frac{3}{\varepsilon}-1.\]

Proposition 1. \(\displaystyle{\lim\left(\frac{3n^2}{n^2+1}\right)=3}\).

Proof. Let \(\varepsilon>0\) be given. By the Archimedean Property there is a natural number \(K(\varepsilon)>0\) such that

\[K(\varepsilon)\geq \sqrt{\frac{3}{\varepsilon}}.\]

Let \(n\in\N\) such that \(n\geq K(\varepsilon)\) be arbitrary. This implies \(n\geq \sqrt{\frac{3}{\varepsilon}}\), and hence \(n^2\geq \frac{3}{\varepsilon}\). This implies

\[n^2+1>n^2\geq \frac{3}{\varepsilon}.\]

Multiplying both sides of this inequality by the positive number \(\frac{\varepsilon}{n^2+1}\) we obtain

\[\varepsilon>\frac{3}{n^2+1}.\]

Finally, we have

\[\left|\frac{3n^2}{n^2+1}-3\right| = \left|\frac{3n^2}{n^2+1}-\frac{3n^2+3}{n^2+1}\right| = \left|\frac{-3}{n^2+1}\right| = \frac{3}{n^2+1}<\varepsilon.\ \Box\]

Prove the following: \(\displaystyle{\lim\left(\frac{n-3}{n+2}\right)=1}\).

Prove the following: \(\displaystyle{\lim\left(\frac{n-3}{n+2}\right)=1}\).