6

Homework 6.pdf

1. Prove using definition that

   1. \(\lim_{x\to2}3x^2-1=11\)

      Use definition, we need \(\forall\varepsilon>0,\exists\delta>0,0<\left|x-2\right|<\delta:\left|3x^2-1-11\right|<\varepsilon\)​

      Since \(3x^2-12=3\left(x^2-4\right)=3\left(x-2\right)\left(x+2\right)\), we already have \(0<|x-2|<\delta\), then we need to bound \(x+2\)​

      Consider \(\delta=1\), then \(1<x<3\Rightarrow3<x+2<5\), thus \(|x+2|<5\)​

      Thus \(\forall\varepsilon>0,\exists\delta=\frac{\varepsilon}{15},0<\left|x-2\right|<\frac{\varepsilon}{15}:\left|3x^2-1-11\right|=3\left|x-2\right|\left|x+2\right|<\varepsilon\)

      Thus \(\lim_{x\to2}3x^2-1=11\) 2. \(\lim_{x \to +\infty}\frac{1}{\sqrt{x}} = 0\)

      Use definition, we have \(\forall\varepsilon>0,\exists N=\frac{1}{\varepsilon^2},\forall x>N:\frac{1}{\sqrt{x}}<\varepsilon\)

      Since \(x\to +\infty\), then \(\frac{1}{\sqrt{x}}=\left|\frac{1}{\sqrt{x}}\right|\), then we have \(\forall\varepsilon>0,\exists N=\frac{1}{\varepsilon^2},\forall x>N:\left|\frac{1}{\sqrt{x}}\right|<\varepsilon\)​

      Thus \(\lim_{x \to +\infty}\frac{1}{\sqrt{x}} = 0\) 3. \(\lim_{x \to 0^+}\frac{1}{\sin x} = +\infty\)

      Use definition, we have \(\forall N>0,\exists\delta=\arcsin\frac{1}{N},x\in\left(0,\delta\right):|\frac{1}{\sin x}|=\frac{1}{\sin x}>\frac{1}{\sin\delta}=\frac{1}{\sin\left(\arcsin\frac{1}{N}\right)}=\frac{1}{\frac{1}{N}}=N\)

      Thus \(\lim_{x \to 0^+}\frac{1}{\sin x} = +\infty\)

2. (Contraction Mapping Theorem). Let \(f\) be a function defined in all \(\mathbb{R}\), and assume that there is a constant \(c\) such that \(0 < c < 1\) and \(|f(x) - f(y)| \leq c|x - y|\) for all \(x, y \in \mathbb{R}\).

   1. Show that \(f\) is continuous on \(\mathbb{R}\).

      Since \(|f(x)-f(y)|\leq c|x-y|\), then let \(y=x_0\), we have \(|f(x)-f(x_0)|\leq c|x-x_0|<c\delta=\varepsilon\)

      Then we take \(\delta=\frac{\varepsilon}{c}\) which is \(\forall \epsilon > 0, \exists \delta > 0\) such that \(|x-x_0|<\delta\implies|f(x)-f(x_0)|\leq c|x-x_0|<c\delta=\varepsilon\)​

      Thus \(f\) is continuous on \(\mathbb{R}\). 2. Pick some point \(y_1 \in \mathbb{R}\) and construct the sequence \((y_1, f(y_1), f(f(y_1)), \dots)\). In general, if \(y_{n+1} = f(y_n)\), show that the resulting sequence \((y_n)\) is a Cauchy sequence. Hence, let \(y = \lim_{n \to \infty} y_n\).

      We need \(\forall\varepsilon>0,\exists n_0,\forall n+1,n+k>n_0:\left|y_{n+k}-y_{n+1}\right|<\varepsilon\)

      Consider \(\left|y_{n+k}-y_{n+1}\right|=|y_{n+k}-y_{n+k-1}+y_{n+k-1}-y_{n+k-2}+\cdots+y_{n+2}-y_{n+1}|\leq|y_{n+k}-y_{n+k-1}\left|+\right|y_{n+k-1}-y_{n+k-2}\left|+\cdots+\right|y_{n+2}-y_{n+1}|\)

      Then we use \(y_{n+1} = f(y_n)\), then \(=|f\left(y_{n+k-1}\right)-f\left(y_{n+k-2}\right)\left|+\right|f\left(y_{n+k-2}\right)-f\left(y_{n+k-3}\right)\left|+\cdots+\right|f\left(y_{n+1}\right)-f\left(y_{n}\right)|\) (*)

      Then, we use \(0 < c < 1\) and \(|f(x) - f(y)| \leq c|x - y|\)

      Let's analyze \(|f\left(y_{n+k-1}\right)-f\left(y_{n+k-2}\right)|\leq c\left|y_{n+k-1}-y_{n+k-2}\right|\leq c|f\left(y_{n+k-2}\right)-f\left(y_{n+k-3}\right)|\leq c^2\left|y_{n+k-2}-y_{n+k-3}\right|\)

      \(\leq\ldots\leq c^{k-1}\left|y_{n+1}-y_{n}\right|\)

      Then back to (*), we have \(\leq c^{k-1}\left|y_{n+1}-y_{n}\right|+\cdots+c\left|y_{n+1}-y_{n}\right|=\frac{c\left(1-c^{k-1}\right)}{1-c}\left|y_{n+1}-y_{n}\right|\leq\frac{c}{1-c}\left|y_{n+1}-y_{n}\right|\)

      Similarly, let's analyze \(|y_{n+1}-y_{n}|=|f\left(y_{n}\right)-f\left(y_{n-1}\right)|\leq c|y_{n}-y_{n-1}|\leq\ldots\leq c^{n-1}|y_2-y_1|\)​

      Finally, \(\left|y_{n+k}-y_{n+1}\right|\leq|y_2-y_1|\frac{c^{n}}{1-c}\leq|y_2-y_1|\frac{c^{n_0}}{1-c}\), we can choose \(n_0=\left\lbrack\log_{c}\frac{\varepsilon\left(1-c\right)}{\left|y_2-y_1\right|}\right\rbrack+1\)​

      Which is \(\forall\varepsilon>0,\exists n_0=\left\lbrack\log_{c}\frac{\varepsilon\left(1-c\right)}{\left|y_2-y_1\right|}\right\rbrack+1,\forall n+1,n+k>n_0:\left|y_{n+k}-y_{n+1}\right|<\varepsilon\)

      Thus \((y_n)\) is a Cauchy sequence.

      Since Cauchy sequence is convergent, then \(y = \lim_{n \to \infty} y_n\). 3. Prove that \(y\) is a fixed point, that is \(f(y) = y\).

      We need to prove \(f\left(\lim_{n\to\infty}y_{n}\right)=y\), since \(f\) is continuous, then we just to prove \(\lim_{n\to\infty}f\left(y_{n}\right)=y\)

      Since \((y_n)\) has a limit, then \(\lim_{n\to\infty}y_{n}=\lim_{n\to\infty}y_{n+1}\)

      Since \(f(y_n)=y_{n+1}\), then \(\lim_{n\to\infty}y_{n}=\lim_{n\to\infty}f\left(y_{n}\right)\Rightarrow y=\lim_{n\to\infty}f\left(y_{n}\right)\)​

      Thus \(y=\lim_{n\to\infty}f\left(y_{n}\right)\), then \(f\left(\lim_{n\to\infty}y_{n}\right)=y\).

      Thus \(f(y) = y\)

3. Consider the function \(f(x)=\begin{cases}5a^2x-9 & \text{if }x>3,\\ 9\sqrt{a} & \text{if }x=3,\\ \frac{2x^2a}{3} & \text{if }x<3.\end{cases}\)​

   Determine all values of \(a\) such that \(\lim_{x \to 3} f(x)\) exists.

   The domain of \(a\) is \((0,\infty)\)​

   Then \(\lim_{x\to3^{+}}f(x)=\lim_{x\to3^{+}}5a^2x-9=15a^2-9\)

   \(\lim_{x\to3^{-}}f(x)=\lim_{x\to3^{-}}\frac{2x^2a}{3}=\frac{18a}{3}=6a\)​

   Then if \(\lim_{x \to 3} f(x)\) exists, then \(\lim_{x\to3^{+}}f(x)=\lim_{x\to3^{-}}f(x)\Rightarrow15a^2-9=6a\Rightarrow a=1,-\frac35\)

   But \(a>0\), then \(a=1\)

4. Compute the following limits:

   1. \(\lim_{x \to -3} \frac{x^2 + 6x + 9}{\sin(x^2 - 9)}\)

      \(\lim_{x\to-3}\frac{x^2+6x+9}{\sin(x^2-9)}=\lim_{x\to-3}\left(\frac{x^2-9}{\sin(x^2-9)}+\frac{6x+18}{\sin(x^2-9)}\right)=1+6\lim_{x\to-3}\frac{x+3}{\sin(x^2-9)}\)

      \(\lim_{x\to-3}\frac{x+3}{\sin(x^2-9)}=\lim_{x\to-3}\left(\frac{\left(x-3\right)\left(x+3\right)}{\sin(x^2-9)}\cdot\frac{1}{x-3}\right)=\lim_{x\to-3}\frac{1}{x-3}=-\frac16\)​

      Thus \(\lim_{x \to -3} \frac{x^2 + 6x + 9}{\sin(x^2 - 9)}=0\) 2. \(\lim_{x \to 1} \frac{x^m - 1}{x^n - 1}, \, n,m \in \mathbb{N}\)

      \(\lim_{x\to1}\frac{x^{m}-1}{x^{n}-1}=\lim_{x\to1}\frac{\left(x-1\right)\left(x^{m-1}+x^{m-2}+\cdots+1\right)}{\left(x-1\right)\left(x^{n-1}+x^{n-2}+\cdots+1\right)}=\lim_{x\to1}\frac{\left(x^{m-1}+x^{m-2}+\cdots+1\right)}{\left(x^{n-1}+x^{n-2}+\cdots+1\right)}=\frac{m}{n}\) 3. \(\lim_{x\to+\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}\)

      Consider \(\frac{\sqrt{x}}{\sqrt{x}+1}\leq\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}\leq\frac{\sqrt{x+\sqrt{x+\left(\sqrt{x}\right)^2}}}{\sqrt{x}}=\frac{\sqrt{x+\sqrt{2x}}}{\sqrt{x}}\)

      \(\lim_{x\to+\infty}\frac{\sqrt{x}}{\sqrt{x}+1}=\lim_{x\to+\infty}\frac{1}{1+\frac{1}{\sqrt{x}}}=1\)

      \(\lim_{n\to+\infty}\frac{\sqrt{x+\sqrt{2x}}}{\sqrt{x}}=\sqrt{\lim_{n\to+\infty}\frac{x+\sqrt{2x}}{x}}=\sqrt{\lim_{n\to+\infty}\frac{1+\sqrt{\frac{2}{x}}}{1}}=1\)​

      By sandwich theorem, \(\lim_{x\to+\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{\sqrt{x+1}}=1\)

5. Decide if the following sentences are True or False:

   1. If \(|f|\) is continuous in \(x_0\) then \(f\) is continuous in \(x_0\).

      False, counter example: \(f(x)=\begin{cases}-1\text{ if }x<1\\1\text{ if }x\geq1\end{cases}\)​

      \(|f|\) is continuous in \(x_0=1\) but \(f\) is not continuous in \(1\). 2. If \(f(x)\) and \(g(x) \neq 0\) for all \(x\) and \(f^2(x) = g^2(x)\), then \(f = g\) or \(f = -g\).

      True

      Since \(f(x),g(x)\neq 0\), then \(\frac{f^2\left(x\right)^{}}{g^2\left(x\right)}=1\Rightarrow\frac{f\left(x\right)^{}}{g\left(x\right)}=\pm1\Rightarrow f=\pm g\)​ 3. There is a function \(f(x)\) which is not continuous at \(0\) but \(f(x) + \frac{1}{f(x)}\) is continuous at \(0\).

      True

      Since \(f(x)\) which is not continuous at \(0\), then \(\lim_{x\to0^{+}}f\left(x\right)=a\neq b=\lim_{x\to0^{-}}f\left(x\right)\)​

      But \(f(x) + \frac{1}{f(x)}\) is continuous at \(0\), then \(\lim_{x\to0^{+}}\left(f(x)+\frac{1}{f(x)}\right)=\lim_{x\to0^{-}}\left(f(x)+\frac{1}{f(x)}\right)\)​

      Thus \(a+\frac{1}{a}=b+\frac{1}{b}\Rightarrow a-b=\frac{1}{b}-\frac{1}{a}\Rightarrow a-b=\frac{a-b}{ab}\Rightarrow ab=1\)​

      Thus it is possible

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