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11.1 Limit

Tutorial Week 3 part 2.pdf

  1. (a) \(\lim_{n\to\infty}\left(n-\sqrt{n^2+2n}\right)=\lim_{n\to\infty}\frac{\left(n-\sqrt{n^2+2n}\right)\left(n+\sqrt{n^2+2n}\right)}{n+\sqrt{n^2+2n}}=\lim_{n\to\infty}\frac{n^2-\left(\sqrt{n^2+2n}\right)^2}{n+\sqrt{n^2+2n}}\)

    \(=\lim_{n\to\infty}\frac{n^2-(n^2+2n)}{n+\sqrt{n^2+2n}}=\lim_{n\to\infty}\frac{-2n}{n+\sqrt{n^2+2n}}=\lim_{n\to\infty}\frac{-2n}{n+\sqrt{n^2\left(1+\frac{2}{n}\right)}}\)

    \(=\lim_{n\to\infty}\frac{-2n}{n+\sqrt{n^2}\cdot\sqrt{1+\frac{2}{n}}}=\lim_{n\to\infty}\frac{-2n}{n+n\sqrt{1+\frac{2}{n}}}=\lim_{n\to\infty}\frac{-2}{1+\sqrt{1+\frac{2}{n}}}\)

    \(\lim_{n\to\infty}\left(1+\sqrt{1+\frac{2}{n}}\right)=\lim_{n\to\infty}1+\lim_{n\to\infty}\sqrt{1+\frac{2}{n}}=1+1=2\)

    \(=\frac{-2}{2}=-1\)

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    (b) \(\lim_{n \to \infty} \sqrt[n]{2 + (-1)^n}\)

    \((-1)^{n}=\begin{cases}1 & \text{if }n\text{ is even}\\ -1 & \text{if }n\text{ is odd}\end{cases}\)

    Since \(1 \leq 2 + (-1)^n \leq 3\) then \(\sqrt[n]{1} \leq \sqrt[n]{2 + (-1)^n} \leq \sqrt[n]{3}\)

    By the Sandwich Theorem:

    \(\lim_{n \to \infty} \sqrt[n]{1} = 1\) and \(\lim_{n \to \infty} \sqrt[n]{3} = 1\) \(\Rightarrow \lim_{n \to \infty} \sqrt[n]{2 + (-1)^n} = 1\)

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    (c) \(\lim_{n \to \infty} \sqrt[n]{2^n + 3^n} = 3\)

    Let \(a_{n+1} = 2^{n+1} + 3^{n+1}\)

    Then, \(\frac{a_{n+1}}{a_n} = \frac{2^{n+1} + 3^{n+1}}{2^n + 3^n} = \frac{3^n \left( \frac{2^{n+1}}{3^n} + 3 \right)}{3^n \left( \frac{2^n}{3^n} + 1 \right)}\) \(=\frac{\left(\frac23\right)^{n}\cdot2+3}{\left(\frac23\right)^{n}+1}\)

    Thus, \(\lim_{n\to\infty}\frac{a_{n+1}}{a_{n}}=\lim_{n\to\infty}\frac{\left(\frac23\right)^{n}\cdot2+3}{\left(\frac23\right)^{n}+1}\)

    Since \(\lim_{n \to \infty} \left(\frac{2}{3}\right)^n = 0\), we have \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \frac{3}{1} = 3\)

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    Theorem: Let \(a_n\), \(a_n > 0\), if \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L\) \(\Rightarrow \lim_{n \to \infty} \sqrt[n]{a_n} = L\)

    Corollary: \(\lim_{n \to \infty} \sqrt[n]{n} = 1\)

    Proof: Let \(a_n = n+1\)

    \(\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = \lim_{n \to \infty} \frac{n+1}{n} = 1\) \(\Rightarrow \lim_{n \to \infty} \sqrt[n]{n} = 1\)

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    (d) \(\lim_{n \to \infty} \frac{\sqrt[n]{1 + 2^n + 3^n + \dots + n^n}}{n}\)

    Since \(n^{n}\leq1+2^{n}+3^{n}+\dots+n^{n}\leq n\cdot n^{n}\) for large \(n\),

    \(1=\frac{\sqrt[n]{n^{n}}}{n}\leq\frac{\sqrt[n]{1+2^{n}+3^{n}+\dots+n^{n}}}{n}\leq\frac{\sqrt[n]{n\cdot n^{n}}}{n}\)

    Thus, applying the Sandwich Theorem:

    \(\lim_{n \to \infty} \frac{\sqrt[n]{1 + 2^n + 3^n + \dots + n^n}}{n} = \lim_{n \to \infty} \frac{\sqrt[n]{n \cdot n^n}}{n}\) \(=\lim_{n\to\infty}\sqrt[n]{n}=1\)

  2. Let \(A\) be a non-empty bounded set and \(s=\sup A\). Prove that there exists a sequence \((a_n)\) such that \(a_n \in A\) \(\forall n\) and \(\lim_{n \to \infty} a_n = s\)

    Analysis: Notice that the problem asks you to prove the existence, not the limit is s

    So you can choose any \(\varepsilon\) as you want because we see \(a_n \in A\) \(\forall n\) and \(\lim_{n \to \infty} a_n = s\) as default

    How to prove existence? We need to prove \(a_n\) in the \((s-\varepsilon,s)\) for any \(\varepsilon\) and \(a_n\) should get close to \(s\) as \(n\) grows

    Thus we need construct a sequence of \(\varepsilon\) which is decreasing

    Thus for example \(a_1\in\left(s-0.3,s\right)\), \(a_2\in(s-0.2,s)\), \(a_3\in(s-0.1,s)\)...

    Main idea: Put \(a_n\) is a smaller and smaller interval

    Proof

    \(s = \sup A \Rightarrow \forall \ \epsilon > 0, \ \exists \ a \in A\) such that \(s - \epsilon < a < s\)

    For example, if \(\epsilon = 1\), then there exists \(a_1 \in A\) such that \(s - 1 < a_1 < s\)

    If \(\epsilon = \frac{1}{2}\), there exists \(a_2 \in A\) such that \(s - \frac{1}{2} < a_2 < s\)

    If \(\epsilon = \frac{1}{3}\), there exists \(a_3 \in A\) such that \(s - \frac{1}{3} < a_3 < s\)

    ...

    For \(\epsilon = \frac{1}{n}\), there exists \(a_n \in A\) such that \(s - \frac{1}{n} < a_n < s\)

    image​

    Let \((a_n) = (a_1, a_2, \dots)\), now let us prove that \(\lim_{n \to \infty} a_n = s\)

    Let \(C > 0\). We need to show that \(|a_n - s| < C\) for all \(n \geq N\) for some \(N\).

    Since \(|a_n - s| < \frac{1}{n} < C\), we can choose \(N > \frac{1}{C}\)

    Then, with this choice of \(N\), we have that \(\lim_{n \to \infty} a_n = s\)

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