# -*- fill-column: 80; org-confirm-babel-evaluate: nil -*- #+TITLE: Assignment 14, Infinitesimal Calculus #+AUTHOR: Oleg Sivokon #+EMAIL: olegsivokon@gmail.com #+DATE: <2015-04-03 Fri> #+DESCRIPTION: Fourth asssignment in the course Infinitesimal Calculus #+KEYWORDS: Infinitesimal Calculus, Assignment, Limits of functions #+LANGUAGE: en #+LaTeX_CLASS: article #+LATEX_CLASS_OPTIONS: [a4paper] #+LATEX_HEADER: \usepackage[usenames,dvipsnames]{color} #+LATEX_HEADER: \usepackage{commath} #+LATEX_HEADER: \usepackage{tikz} #+LATEX_HEADER: \usetikzlibrary{shapes,backgrounds} #+LATEX_HEADER: \usepackage{marginnote} #+LATEX_HEADER: \usepackage{listings} #+LATEX_HEADER: \usepackage{color} #+LATEX_HEADER: \usepackage{enumerate} #+LATEX_HEADER: \hypersetup{urlcolor=blue} #+LATEX_HEADER: \hypersetup{colorlinks,urlcolor=blue} #+LATEX_HEADER: \setlength{\parskip}{16pt plus 2pt minus 2pt} #+LATEX_HEADER: \definecolor{codebg}{rgb}{0.96,0.99,0.8} #+LATEX_HEADER: \definecolor{codestr}{rgb}{0.46,0.09,0.2} #+LATEX_HEADER: \DeclareMathOperator{\Dom}{Dom} #+LATEX_HEADER: \allowdisplaybreaks[4] #+BEGIN_SRC emacs-lisp :exports none (setq org-latex-pdf-process '("latexmk -pdflatex='pdflatex -shell-escape -interaction nonstopmode' -pdf -f %f") org-latex-listings t org-src-fontify-natively t org-listings-escape-inside '("(*@" . "@*)") org-latex-custom-lang-environments '((maxima "maxima")) org-babel-latex-htlatex "htlatex") (defmacro by-backend (&rest body) `(cl-case (when (boundp 'backend) (org-export-backend-name backend)) ,@body)) (defmacro with-current-dir (directory &rest body) `(let ((old default-directory)) (cd ,directory) (unwind-protect (progn ,@body) (cd old)))) (defun format-maxima-graph (maxima-output) (let ((tex (replace-regexp-in-string "[] ]+$" "" (replace-regexp-in-string "^\\s-+" "" (car (reverse (split-string maxima-output "\n"))))))) (with-current-dir (format "%s/images/" (file-name-directory (buffer-file-name))) (shell-command (format "latexmk -pdflatex='pdflatex -shell-escape -interaction nonstopmode' -pdf -f %s" tex))) (format "%s.pdf" (file-name-sans-extension tex)))) #+END_SRC #+RESULTS: : format-maxima-graph #+NAME: fname #+HEADER: :var f="dummy" #+BEGIN_SRC emacs-lisp :exports none (format "\"%s/images/%s\"" (file-name-directory (buffer-file-name)) f) #+END_SRC #+BEGIN_LATEX \definecolor{codebg}{rgb}{0.96,0.99,0.8} \lstnewenvironment{maxima}{% \lstset{backgroundcolor=\color{codebg}, escapeinside={(*@}{@*)}, aboveskip=20pt, showstringspaces=false, frame=single, framerule=0pt, basicstyle=\ttfamily\scriptsize, columns=fixed}}{} } \makeatletter \newcommand{\verbatimfont}[1]{\renewcommand{\verbatim@font}{\ttfamily#1}} \makeatother \verbatimfont{\small}% \clearpage #+END_LATEX * Problems ** Problem 1 1. Find the domain of $f$ defined as $f(x) = \sqrt{\tan x - 1}$. 2. Find all values of $x$ in segment $[0, \pi]$, for which $\abs{\tan x} \leq \sin 2x}$. *** Answer 1 $\tan x > 1$ where $\frac{\pi}{4} < x \bmod \frac{\pi}{2} < \frac{\pi}{2}$. Since we assume that $f$ is real-valued, we cannot extract roots from negative numbers. Hence $\Dom(f) = \{x \in \mathbb{R} \; | \; \frac{\pi}{4} < x \bmod \frac{\pi}{2} < \frac{\pi}{2}\}$. #+NAME: prob1 #+HEADER: :exports source #+HEADER: :var %out=fname(f="tangent_x.tex") #+BEGIN_SRC maxima :results output raw programmode: false; print (plot2d (tan(x), [x, -2 * %pi, 2 * %pi], [y, -3, 3], [gnuplot_pdf_term_command, "set term cairolatex standalone pdf size 16cm,10.5cm"], [gnuplot_postamble, "set arrow from pi/4,-3 to pi/4,3 nohead set arrow from pi*5/4,-3 to pi*5/4,3 nohead set arrow from -2*pi,1 to 2*pi,1 nohead"], [ytics, 1, 1, 1], [xtics, 0, 1, 0], [label, ["${\\displaystyle \\frac{\\pi}{4}}$", 1.4, 2.4], ["${\\displaystyle \\frac{5\\pi}{4}}$", 3.2, 2.4]], [title, "Tangent curve"], [pdf_file, %out])); #+END_SRC #+HEADER: :results raw file :exports results #+BEGIN_SRC emacs-lisp :var maxima-output=prob1(%out=fname(f="tangent_x.tex")) (format-maxima-graph maxima-output) #+END_SRC #+ATTR_LATEX: :width 0.9\textwidth #+RESULTS: [[file:/home/wvxvw/Documents/uni/infinitesimal-calculus//images/tangent_x.pdf]] *** Answer 2 Both functions attain the same values at 0 and $\frac{\pi}{4}$. But sine is a concave function and tanget is a convex function, thus tangent must be less than sine at this interval. Tangent keeps increasing until $\frac{\pi}{2}$, while sine will be decreasing until $\frac{3\pi}{2}$, thus, on this interval tangent is greater than sine. The functions meet again at $x=\pi$. #+NAME: prob2 #+HEADER: :exports source #+HEADER: :var %out=fname(f="abstangent_x.tex") #+BEGIN_SRC maxima :results output raw programmode: false; gnuplot_pdf_command: %command; print(plot2d ([abs(tan(x)), sin(2 * x)], [x, -2 * %pi, 2 * %pi], [y, -1.5, 3], [gnuplot_pdf_term_command, "set term cairolatex standalone pdf size 16cm,10.5cm"], [gnuplot_postamble, "set arrow from pi/4,-1.5 to pi/4,3 nohead set arrow from pi,-1.5 to pi,3 nohead set arrow from -2*pi,1 to 2*pi,1 nohead set key spacing 1.8 left bottom"], [xtics, 0, 1, 0], [ytics, 1, 1, 1], [label, ["${\\displaystyle \\frac{\\pi}{4}}$", 1.4, 2.1], ["${\\displaystyle \\pi}$", 3.4, 2.1]], [title, "Tangent curve intersecting with sine curve"], [pdf_file, %out])); #+END_SRC #+HEADER: :results raw file :exports results #+BEGIN_SRC emacs-lisp :var maxima-output=prob2(%out=fname(f="abstangent_x.tex")) (format-maxima-graph maxima-output) #+END_SRC #+ATTR_LATEX: :width 0.9\textwidth #+RESULTS: [[file:/home/wvxvw/Documents/uni/infinitesimal-calculus//images/abstangent_x.pdf]] ** Problem 2 Let $f$, $g$ and $h$ be functions from $\mathbb{R}$ to $\mathbb{R}$. 1. If $f \circ g = f \circ h$, does it follow $g = h$? 2. If $f \circ g = f \circ h$, and $f$ is one-to-one, does it follow $g = h$? 3. If $f \circ g = f \circ h$, and $f$ is onto, does it follow $g = h$? 4. If $f \circ g$ is increasing, and $f$ is decreasing, does it follow that $g$ is increasing? 5. If $f \circ g$ is increasing, and $f$ is one-to-one, does it follow that $g$ is monotonic? *** Answer 3 No, $g$ and $h$ are not necessarily equal. Whenever co-domain of $f$ doesn't contain some real number, $g$ and $h$ may differ in that input. For example, let $f(x) = 2x$, $g(x) = (x \bmod 2) + x$ and $h(x) = (x \bmod 2) * 2 + x$. Because $f$ in this example will only generate even numbers, the $(x \bmod 2)$ term will always be zero, thus $f \circ g = f \circ h$, but, obviously, $g \neq h$. *** Answer 4 No, it isn't sufficient for $f$ to be one-to-one to ensure right-cancellation property under composition. The example given in [[Answer 3]] is applicable in this case too since whenever $f(x) = f(y)$ so is $x = y$ (since multiplication does have the cancellation property). *** Answer 5 Yes, if $f$ is onto, then the composition is right-cancellable. Suppose, for contradiction it wasn't, then for some $y$ $g(y) \neq h(y)$, but $y = f(x)$ (since by definition of a total function, every element in its co-domain has an element in its domain). Hence $g(f(x)) \neq h(f(x))$, but we are given that $g \circ f = h \circ f$, which is a contradiction. Hence functions are equal. *** Answer 6 No, $g$ doesn't need to be increasing. Put $g(x) = f(x) = -x$, both $f$ and $g$ are decreasing but $f \circ g = Id$, which is an increasing function. *** Anser 7 No, $g$ is not necessarily monotonic. Put $f(x) = x(-1)^x$ and $g(x) = \abs{x}$. Then $(f \circ g)(x) = \abs{x(-1)^x} = x\abs{(-1)^x} = x$. $f \circ g$ is increasing, $f$ is one-to-one, but $g$ isn't monotonic: it decreases whenever $x$ is negative and increases whenever $x$ is positive. ** Problem 3 1. Prove from $\epsilon-\delta$ definition of limit that $\lim_{x \to 2}\sqrt{3x - 2} = 2$. 2. Prove from $\epsilon-M$ definition of limit that $\lim_{x \to \infty}\frac{x}{x+\sin x} = 1$. *** Answer 8 Recall the definition: #+BEGIN_QUOTE For all $\epsilon > 0$ there exists $\sigma > 0$ s.t. for all $x$ in $\Dom(f(x))$ which satisfy $0 < \abs{x - x_0} < \sigma$ the inequality $\abs{f(x) - L} < \epsilon$ holds. #+END_QUOTE Let $\epsilon$ be arbitrary real, put #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{equation*} \begin{aligned} &\abs{f(x) - L} &< \epsilon &\iff \\ &\abs{\sqrt{3x - 2} - 2} &< \epsilon \\ &\textit{Suppose $\sqrt{3x - 2} - 2 > 0$} \\ 0 &< \sqrt{3x - 2} - 2 &< \epsilon &\iff \\ 0 &< \sqrt{3x - 2} &< \epsilon + 2 &\iff \\ 0 &< 3x - 2 &< (\epsilon + 2)^2 &\iff \\ 0 &< 3x - 2 &< \epsilon^2 + 4\epsilon + 4 &\iff \\ 0 &< 3x - 6 &< \epsilon^2 + 4\epsilon &\iff \\ 0 &< x - 2 &< \frac{\epsilon^2 + 4\epsilon}{3} \\ &\textit{Similarly for $\sqrt{3x - 2} - 2 < 0$} \\ 0 &> \sqrt{3x - 2} - 2 &> -\epsilon &\iff \\ 0 &> \sqrt{3x - 2} &> -\epsilon + 2 &\iff \\ 0 &> 3x - 2 &> (-\epsilon + 2)^2 &\iff \\ 0 &> 3x - 2 &> \epsilon^2 - 4\epsilon + 4 &\iff \\ 0 &> 3x - 6 &> \epsilon^2 - 4\epsilon &\iff \\ 0 &> x - 2 &> \frac{\epsilon^2 - 4\epsilon}{3} \end{aligned} \end{equation*} #+END_SRC Hence, we can choose $\delta$ to be $\frac{\epsilon^2 - 4\epsilon}{3}$ whenever $x < x_0$ and $\frac{\epsilon^2 + 4\epsilon}{3}$ whenever $x > x_0$, which completes the proof. *** Answer 9 Recall the definition: #+BEGIN_QUOTE For all $\epsilon > 0$ there exists $M > 0$ s.t. for all $x$ in $\Dom(f(x))$ $x > M$ implies $\abs{f(x) - L} < \epsilon$. #+END_QUOTE Let $\epsilon > 0$, then look for appropriate value for $x$: #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{align*} &\abs{\frac{x}{x + \sin x} - 1} &< \epsilon &\iff \\ &\abs{\frac{x}{x + \sin x} - \sin^2 x - \cos^2 x} &< \epsilon &\iff \\ &\abs{\frac{x - \sin^2 x(x + \sin x) - \cos^2 x(x + \sin x)}{x + \sin x}} &< \epsilon &\iff \\ &\abs{\frac{x - x\sin^2 x - \sin^3 x - x\cos^2 x - \cos^2 x \sin x}{x + \sin x}} &< \epsilon &\iff \\ &\abs{\frac{x(1 - \sin^2 - \cos^2 x) - \sin x(\sin^2 x - \cos^2)}{x + \sin x}} &< \epsilon &\iff \\ &\abs{\frac{x(1 - 1) - \sin x(1)}{x + \sin x}} &< \epsilon &\iff \\ &\abs{\frac{-\sin x}{x + \sin x}} &< \epsilon \\ \end{align*} #+END_SRC Assume $x > 0$: #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{align*} &\frac{-\sin x}{x + \sin x} &< \epsilon &\iff \\ &-\sin x &< \epsilon(x + \sin x) &\iff \\ &-\sin x &< \epsilon x + \epsilon \sin x &\iff \\ &-\epsilon x &< \sin x + \epsilon \sin x &\iff \\ &-x &< \frac{\sin x + \epsilon \sin x}{\epsilon} &\iff \\ &x &> \frac{\sin x(1 + \epsilon)}{\epsilon} \end{align*} #+END_SRC Put $M = \max\left(0, \frac{\sin x(1 + \epsilon)}{\epsilon}\right)$. If $x > M$, then $x > 0$ and $x > \frac{\sin x(1 + \epsilon)}{\epsilon}$, hence: #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{align*} &x &> \frac{\sin x(1 + \epsilon)}{\epsilon} \\ &\hdots \textit{Reverse the calculations above} \\ &\frac{-\sin x}{x + \sin x} &< \epsilon \\ &\hdots \\ &\abs{\frac{x}{x + \sin x} - 1} &< \epsilon. \end{align*} #+END_SRC Which completes the proof. ** Problem 4 1. Let $f$ be a function defined in the neighborhood of $x_0$. Express ``$f$ doesn't have a limit at $x_0$'' using: + $\epsilon-\sigma$ language. + Using Heine definition of limit (for sequences). 2. Prove that $f(x) = \frac{x}{x - \lfloor x \rfloor}$ doesn't have a finite limit at $x \to 0$ in the following ways: + By using $\epsilon-\sigma$ definition given above. + By using Heine definition of limit (also given above). *** Answer 10 Recall the $\epsilon-\delta$ definition: #+BEGIN_LATEX \emph{The limit of $f(x)$ at $x_0$ is defined to be $L$ s.t. for every $\epsilon > \abs{f(x) - L} > 0$ we can find $\delta > \abs{x - x_0} > 0$.} #+END_LATEX To negate this is to say that there exists such $\epsilon$ for which we can't find a positive $\delta$ larger than the distance from $x$ to $x_0$. #+BEGIN_LATEX \emph{Heine defines limit to be $L$ whenever for every sequence $(x_n)$ which convergest to $x_0$, every sequence of function values $f(x_n)$ converges to $L$.} #+END_LATEX To negate this definition we claim that there exists a sequence $(x_m)$, which convergest to $x_0$, however $f(x_m)$ doesn't converge to $L$. *** Answer 11 I'll do the Heine first, because it's easier. We can choose sequences $(x_n) = -\frac{\pi}{2nx}$ and $(x_m) = \frac{\pi}{2mx}$, both are immediately reducible to the limit of a fraction as $x$ approaches zero, hence, the limit for both is zero. Now, if we plug them back into $f$, we get: #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{align*} \lim_{x \to 0}\frac{x}{x - \lfloor \sin x \rfloor} &= \lim_{x \to 0}\frac{x}{x - 0} \\ &\textit{$\lfloor \sin x \rfloor = 0$ whenever $0 < x < \frac{\pi}{2}$} \\ &= 1\;. \end{align*} #+END_SRC Similarly: #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{align*} \lim_{x \to 0}\frac{x}{x - \lfloor \sin x \rfloor} &= \lim_{x \to 0}\frac{x}{x + 1} \\ &\textit{$\lfloor \sin x \rfloor = -1$ whenever $0 > x > -\frac{\pi}{2}$} \\ &= 0\;. \end{align*} #+END_SRC Now, the $\epsilon-\sigma$ approach: Since we can pick arbitrary $\epsilon$, put $\epsilon = \frac{1}{2}$. Now, we could try to find the limit withing $\pi$ distance from zero. Assuming thus $x \in (\pi, -\pi)$. #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{align*} -&\frac{1}{2} < \frac{\sigma}{\sigma - \lfloor \sin \sigma \rfloor} - L < \frac{1}{2} \\ -&\frac{1}{2} < \frac{\sigma}{\sigma - 0} - L < \frac{1}{2} \\ -&\frac{1}{2} < 1 - L < \frac{1}{2} \\ -&\frac{3}{2} < -L < -\frac{1}{2} \\ &\frac{3}{2} > L > \frac{1}{2} \\ &\textit{Similarly:} \\ -&\frac{1}{2} < \frac{-\sigma}{\lfloor - \sin \sigma \rfloor -\sigma} - L < \frac{1}{2}\\ -&\frac{1}{2} < \frac{-\sigma}{-\sigma + 1} - L < \frac{1}{2} \\ -&\frac{1}{2} - \frac{-\sigma}{-\sigma + 1} < -L < \frac{1}{2} - \frac{-\sigma}{-\sigma + 1} \\ -&\frac{-\sigma + 1 + 2\sigma}{-2\sigma + 2} < -L < \frac{1}{2} - \frac{-\sigma}{1 - \sigma} \\ -&\frac{1 + \sigma}{-2\sigma + 2} < -L < \frac{1}{2} - \frac{-\sigma}{1 - \sigma} \\ &\frac{-1 - \sigma}{-2\sigma + 2} < -L < \frac{1}{2} - \frac{-\sigma}{1 - \sigma} \\ &\frac{-1(1 + \sigma)}{-1(2\sigma - 2)} < -L < \frac{1}{2} - \frac{-\sigma}{1 - \sigma} \\ &\frac{1 + \sigma}{2\sigma - 2} < -L < \frac{1}{2} - \frac{-\sigma}{1 - \sigma} \\ &\textit{Since $\sigma > 0$} \\ &\frac{1 + \sigma}{2\sigma + 2} < -L < \frac{1}{2} - \frac{-\sigma}{1 - \sigma} \\ &\frac{1}{2} < -L < \frac{1}{2} - \frac{-\sigma}{1 - \sigma} \\ &\textit{Contradiction: $L > \frac{1}{2}$ and $L < \frac{1}{2}$.} \end{align*} #+END_SRC #+NAME: prob3 #+HEADER: :exports source #+HEADER: :var %out=fname(f="nolimit.tex") #+BEGIN_SRC maxima :results output raw programmode: false; gnuplot_pdf_command: %command; print(plot2d (x / (x - floor(sin(x))), [x, -2 * %pi, 2 * %pi], [y, -2, 2], [gnuplot_pdf_term_command, "set term cairolatex standalone pdf size 16cm,10.5cm"], [ylabel, "${\\displaystyle \\frac{x}{x - \\lfloor \\sin x \\rfloor}}$"], [title, concat("${\\displaystyle \\lim_{x \\to 0}", "\\frac{x}{x - \\lfloor \\sin x \\rfloor}}$")], [pdf_file, %out])); #+END_SRC #+HEADER: :results raw file :exports results #+BEGIN_SRC emacs-lisp :var maxima-output=prob3(%out=fname(f="nolimit.tex")) (format-maxima-graph maxima-output) #+END_SRC #+RESULTS: [[file:/home/wvxvw/Documents/uni/infinitesimal-calculus//images/nolimit.pdf]] ** Problem 5 Find limits of: 1. $\lim_{x \to 0}\frac{1 - \cos x}{x \sin x}$. 2. $\lim_{x \to 0}\frac{x + 7x^3}{x^3 - 2x^4}$. 3. $\lim_{x \to 0}\frac{x^2 - 1}{2x^3 - x^2 - x}$. 4. $\lim_{x \to 0}(\sqrt{1 + x + x^2} - \sqrt{1 - x + x^2})$. 5. $\lim_{x \to k}\lfloor x \rfloor \tan \frac{\pi x}{2}$, $k = 0, 1, 2$. *** Answer 12 #+NAME: prob3 #+HEADER: :exports both #+BEGIN_SRC maxima :results output raw tex(limit((1 - cos(x)) / (x * sin(x)), x, 0)); #+END_SRC *Proof:* #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{align*} \lim_{x \to 0}\frac{1 - \cos x}{x \sin x} &= \lim_{x \to 0}\frac{(1 - \cos x) * (1 + \cos x)}{x \sin x (1 + \cos x)} \\ &= \lim_{x \to 0}\frac{1 - \cos^2 x}{x \sin x (1 + \cos x)} \\ &= \lim_{x \to 0}\frac{\sin^2 x}{x \sin x (1 + \cos x)} \\ &= \lim_{x \to 0}\frac{\sin x}{x (1 + \cos x)} \\ &= \lim_{x \to 0}\frac{\sin x}{x} * \lim_{x \to 0}\frac{1}{1 + \cos x} \\ &= 1 * \lim_{x \to 0}\frac{1}{1 + \cos x} \\ &= \lim_{x \to 0}\frac{1}{1 + 1} \\ &= \frac{1}{2} \end{align*} #+END_SRC *** Answer 13 #+NAME: prob4 #+HEADER: :exports both #+BEGIN_SRC maxima :results output raw tex(limit((x + 7 * x^3) / (x^3 - 2 * x^4), x, 0)); #+END_SRC *Proof:* #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{align*} \lim_{x \to 0}\frac{x + 7x^3}{x^3 - 2x^4} &= \lim_{x \to 0}\frac{1 + 7x^2}{x^2 - 2x^3} \\ &= \lim_{x \to 0}\frac{1 - 4x^2}{x^2 - 2x^3} + \frac{11x^2}{x^2 - 2x^3} \\ &= \lim_{x \to 0}\frac{(1 - 2x)(1 + 2x)}{x^2(1 - 2x)} + \frac{11}{1 + 2x} \\ &= \lim_{x \to 0}\frac{1 + 2x}{x^2} + \frac{11}{1 + 2x} \\ &= \lim_{x \to 0}\frac{1}{x^2} + \frac{2x}{x^2} + \frac{11}{1 + 2x} \\ &= \lim_{x \to 0}\frac{1}{x^2} + \frac{2}{x} + \frac{11}{1 + 2x} \\ &= \lim_{x \to 0}\frac{1}{x^2} + \lim_{x \to 0}\frac{2}{x} + \lim_{x \to 0}\frac{11}{1 + 2x} \\ &= \infty + \infty + 11 & \textit{Using ifinite limits addition} \\ &= \infty \end{align*} #+END_SRC *** Answer 14 #+NAME: prob5 #+HEADER: :exports both #+BEGIN_SRC maxima :results output raw tex(limit((x^2 - 1) / (2 * x^3 - x^2 - x), x, 0)); #+END_SRC *Proof:* #+HEADER: :exports results #+HEADER: :results (by-backend (pdf "latex") (t "raw")) #+BEGIN_SRC latex \begin{align*} \lim_{x \to 0}\frac{x^2 - 1}{2x^3 - x^2 - x} &= \lim_{x \to 0}\frac{(x - 1)(x + 1)}{x^2(x - 1) + x(x^2 - 1)} \\ &= \lim_{x \to 0}\frac{(x - 1)(x + 1)}{x^2(x - 1) + x(x - 1)(x + 1)} \\ &= \lim_{x \to 0}\frac{(x - 1)(x + 1)}{(x - 1)(x^2 + x(x + 1))} \\ &= \lim_{x \to 0}\frac{x + 1}{x^2 + x(x + 1)} \\ &= \lim_{x \to 0}\frac{x + 1}{x^2 + x^2 + x} \\ &= \lim_{x \to 0}\frac{x + 1}{2x^2 + x} \\ &= \lim_{x \to 0}\frac{x + 1}{x(2x + 1)} \\ &= \lim_{x \to 0}\frac{x + 1}{x} * \lim_{x \to 0}\frac{1}{2x + 1} \\ &= \lim_{x \to 0}\frac{x + 1}{x} * 1 \\ &= \infty \end{align*} #+END_SRC *** Answer 15 #+NAME: prob6 #+HEADER: :exports both #+BEGIN_SRC maxima :results output raw tex(limit(sqrt(1 + x + x^2) - sqrt(1 - x + x^2), x, 0)); #+END_SRC *Proof:* this function is continuous at $x = 0$ since square root is continuous at 1 and this function is offset by one. From definition of continuity we know that the limit of the function coincides with its value, hence the limit is $\sqrt{1 + 0 + 0^2} - \sqrt{1 - 0 + 0^2} = 1 - 1 = 0$. *** Answer 16 #+NAME: prob7 #+HEADER: :exports both #+BEGIN_SRC maxima :results output raw for i : 0 thru 2 do tex(limit(floor(x) * tan((%pi * x) / 2), x, i)); #+END_SRC