From 4d2912426dfb58b6f312118db3aca8277bcba314 Mon Sep 17 00:00:00 2001 From: sean-fitzpatrick Date: Thu, 28 May 2026 08:58:21 -0600 Subject: [PATCH 1/3] Adjust error bounds in limit example --- ptx/sec_limit_def.ptx | 34 +++++++++++++++++++--------------- 1 file changed, 19 insertions(+), 15 deletions(-) diff --git a/ptx/sec_limit_def.ptx b/ptx/sec_limit_def.ptx index f72e247d4..a8910640f 100644 --- a/ptx/sec_limit_def.ptx +++ b/ptx/sec_limit_def.ptx @@ -768,50 +768,54 @@

Since x is approaching 1, we are safe to assume that x is between 0 and 2. + But we need to be careful! The quadratic x^2+x-1 has roots \frac{-1\pm\sqrt{5}}{2}, + and \frac{-1+\sqrt{5}}{2}\approx 0.62 is in the interval [0,2]! + We want to divide by x^2+x-1, so to avoid any division by zero issues, + we choose a smaller interval containing 1: 0.75\lt x\lt 1.25. So - 0\amp \lt x\lt 2 - 0\amp \lt x^2\lt 4\amp\amp \text{(Squared each term.)} - Since 0\lt x\lt 2, we can add 0, - x and 2, respectively, - to each part of the inequality and maintain the inequality. - 0\amp \lt x^2+x\lt 6 - -1\amp \lt x^2+x-1\lt 5\amp\amp \text{(Subtracted 1 from each part.)} + 0\frac34 amp \lt x\lt \frac54 + \frac{9}{16}\amp \lt x^2\lt \frac{25}{16}\amp\amp \text{(Squared each term.)} + Adding the above two inequalities, we get: + \frac{21}{16}\amp \lt x^2+x\lt \frac{45}{16} + \frac{5}{16}\amp \lt x^2+x-1\lt \frac{29}{16}\amp\amp \text{(Subtracted 1 from each part.)}

In Inequality, we wanted \abs{x-1}\lt \varepsilon/\abs{x^2+x-1}. - The above shows that given any x in [0,2], + The above shows that given any x in (0.75,1.25), we know that - x^2+x-1 \amp \lt 5 \amp\amp \text{which implies that} - \frac15 \amp \lt \frac{1}{x^2+x-1} \amp\amp \text{which implies that} - \frac{\varepsilon}5 \amp \lt \frac{\varepsilon}{x^2+x-1}\amp\amp + \abs{x^2+x-1} \amp \lt \frac{29}{16} \amp\amp \text{which implies that} + \frac{16}{25} \amp \lt \frac{1}{\abs{x^2+x-1}} \amp\amp \text{which implies that} + \frac{16\varepsilon}{25} \amp \lt \frac{\varepsilon}{\abs{x^2+x-1}}\amp\amp .

- So we set \delta \lt \epsilon/5. + So we set \delta \lt 16\epsilon/25, on the interval (0.75, 1.25). This ends our scratch-work, and we begin the formal proof (which also helps us understand why this was a good choice of \delta).

- Given \varepsilon, let \delta \lt \varepsilon/5. + Given \varepsilon\gt 0, let \delta \lt \min\{1/4, 16\varepsilon/25\}. We want to show that when \abs{x-1}\lt \delta, then \abs{(x^3-2x)-(-1)}\lt \varepsilon. We start with \abs{x-1}\lt \delta: \abs{x-1} \amp \lt \delta - \abs{x-1} \amp \lt \frac{\varepsilon}{5} + \abs{x-1} \amp \lt \frac{16\varepsilon}{25} \abs{x-1} \amp \lt \frac{\varepsilon}{\abs{x^2+x-1}} \amp\amp \text{(}Inequality, x\text{ near 1)} \abs{x-1}\cdot \abs{x^2+x-1} \amp \lt \varepsilon \abs{x^3-2x+1} \amp \lt \varepsilon \abs{(x^3-2x)-(-1)} \amp \lt \varepsilon , - which is what we wanted to show. + which is what we wanted to show. + (Note that the condition x near 1 is satisfied by the fact that \delta\lt 1/4, + so that 3/4\lt x\lt 5/4.) Thus \lim_{x\to 1}(x^3-2x) = -1.

From 25f9e5ea3ed2640ebcb3d89ce5300ef54cb573c8 Mon Sep 17 00:00:00 2001 From: sean-fitzpatrick Date: Thu, 28 May 2026 13:27:44 -0600 Subject: [PATCH 2/3] add result about continuous function of a sequence --- ptx/sec_sequences.ptx | 6 ++++++ 1 file changed, 6 insertions(+) diff --git a/ptx/sec_sequences.ptx b/ptx/sec_sequences.ptx index 22f3c36d1..3ead51ca2 100644 --- a/ptx/sec_sequences.ptx +++ b/ptx/sec_sequences.ptx @@ -991,6 +991,12 @@ \lim\limits_{n\to\infty} c\cdot a_n = c\cdot L

+ +
  • +

    + If g is continuous at L, then \lim\limits_{n\to\infty}g(a_n)=g(L). +

    +

    • From dfd97a06e492dbe5e838c0054b6a725f2faca5a9 Mon Sep 17 00:00:00 2001 From: sean-fitzpatrick Date: Thu, 28 May 2026 13:57:02 -0600 Subject: [PATCH 3/3] add example involving continuity of piecewise function --- ptx/sec_limit_continuity.ptx | 66 ++++++++++++++++++++++++++++++++++++ 1 file changed, 66 insertions(+) diff --git a/ptx/sec_limit_continuity.ptx b/ptx/sec_limit_continuity.ptx index b8f7cd198..5c2f84d3a 100644 --- a/ptx/sec_limit_continuity.ptx +++ b/ptx/sec_limit_continuity.ptx @@ -754,6 +754,72 @@ + + Checking continuity at a given point + +

    + Let f(x) = \begin{cases}x^2+2x+1,\amp \text{ if } x\lt 0\\2, \amp \text{ if } x=0\\5x+\cos(x^3), \amp \text{ if } x\gt 0\end{cases}. + For each point below, determine whether or not f is continuous at that point, and explain why. +

      +
    1. x=0
      2. +
    2. x=-2
      3. +
    3. x=2
      4. +
    +

    +     + +

    +

      +
    1. +

      + At x=0, we must rely on . + If x\lt 0, then f(x) = x^2+2x+1; therefore, + + \lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(x^2+2x+1)=1 + . + Similarly, for x\gt 0, f(x)=5x+\cos(x^3), so + + \lim_{x\\to 0^+}f(x) = \lim_{x\to 0^+}(5x+\cos(x^3)) = 0+\cos(0)=1 + . + Therefore, \lim_{x\to 1}f(x)=1 exists. +

      + +

      + However, f is not continuous at 0, since + + f(0)=2 \neq 1 = \lim_{x\to 0}f(x) + . +

      +
      2. + +
    2. +

      + If x is near -2, then we can assume that x\lt 0, + and therefore f(x)=x^2+2x+1. + This is a polynomial function, and by , + every polynomial function is continuous on its domain. + Therefore, f is continuous at -2. +

      +
      3. + +
    3. +

      + If x is near 2, then we can assume x\gt 0. + Since 2x and x^3 are polynomial, + we know that these functions are continuous, by . + We know that \cos(x) is continuous by , + and by , the composition of continuous functions is continuous, + so \cos(x^3) is continuous. + Also by , the sum of continuous functions is continuous. + Therefore, for any x\gt 0, f(x)=2x+\cos(x^3) is continuous, and in particular, + f is continuous at x=2. +

      +
      4. +
    +

    +     +     +

    As the video example in illustrates, the above theorems allow us to quickly construct new continuous functions from old ones.