Three small content adjustments

ptx/sec_limit_continuity.ptx

@@ -754,6 +754,72 @@
     </hint>
   </exercise>
 
+    <example xml:id="ex_contpoint">
+    <title>Checking continuity at a given point</title>
+    <statement>
+      <p>
+        Let <m>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}</m>.
+        For each point below, determine whether or not <m>f</m> is continuous at that point, and explain why.
+        <ol marker="(a)">
+          <li><m>x=0</m></li>
+          <li><m>x=-2</m></li>
+          <li><m>x=2</m></li>
+        </ol>
+      </p>
+    </statement>
+    <solution>
+      <p>
+        <ol marker="(a)">
+          <li>
+            <p>
+              At <m>x=0</m>, we must rely on <xref ref="def_continuous"/>.
+              If <m>x\lt 0</m>, then <m>f(x) = x^2+2x+1</m>; therefore,
+              <md>
+                \lim_{x\to 0^-}f(x) = \lim_{x\to 0^-}(x^2+2x+1)=1
+              </md>.
+              Similarly, for <m>x\gt 0</m>, <m>f(x)=5x+\cos(x^3)</m>, so
+              <md>
+                \lim_{x\\to 0^+}f(x) = \lim_{x\to 0^+}(5x+\cos(x^3)) = 0+\cos(0)=1
+              </md>.
+              Therefore, <m>\lim_{x\to 1}f(x)=1</m> exists.
+            </p>
+
+            <p>
+              However, <m>f</m> is not continuous at <m>0</m>, since
+              <md>
+                f(0)=2 \neq 1 = \lim_{x\to 0}f(x)
+              </md>.              
+            </p>
+          </li>
+
+          <li>
+            <p>
+              If <m>x</m> is near <m>-2</m>, then we can assume that <m>x\lt 0</m>,
+              and therefore <m>f(x)=x^2+2x+1</m>.
+              This is a polynomial function, and by <xref ref="thm_poly_rat"/>,
+              every polynomial function is continuous on its domain.
+              Therefore, <m>f</m> is continuous at <m>-2</m>.
+            </p>
+          </li>
+
+          <li>
+            <p>
+              If <m>x</m> is near <m>2</m>, then we can assume <m>x\gt 0</m>.
+              Since <m>2x</m> and <m>x^3</m> are polynomial,
+              we know that these functions are continuous, by <xref ref="thm_poly_rat"/>.
+              We know that <m>\cos(x)</m> is continuous by <xref ref="thm_continuous_functions"/>,
+              and by <xref ref="thm_continuity_algebra"/>, the composition of continuous functions is continuous,
+              so <m>\cos(x^3)</m> is continuous.
+              Also by <xref ref="thm_continuity_algebra"/>, the sum of continuous functions is continuous.
+              Therefore, for any <m>x\gt 0</m>, <m>f(x)=2x+\cos(x^3)</m> is continuous, and in particular, 
+              <m>f</m> is continuous at <m>x=2</m>. 
+            </p>
+          </li>
+        </ol>
+      </p>
+    </solution>
+  </example>
+
   <p xml:id="vidint_limit_con_using_prop" component="video">
     As the video example in <xref ref="vid_limit_con_using_prop"/> illustrates,
     the above theorems allow us to quickly construct new continuous functions from old ones.

ptx/sec_limit_def.ptx

@@ -768,50 +768,54 @@
       <p>
         Since <m>x</m> is approaching <m>1</m>,
         we are safe to assume that <m>x</m> is between <m>0</m> and <m>2</m>.
+        But we need to be careful! The quadratic <m>x^2+x-1</m> has roots <m>\frac{-1\pm\sqrt{5}}{2}</m>,
+        and <m>\frac{-1+\sqrt{5}}{2}\approx 0.62</m> is in the interval <m>[0,2]</m>!
+        We want to divide by <m>x^2+x-1</m>, so to avoid any division by zero issues,
+        we choose a smaller interval containing <m>1</m>: <m>0.75\lt x\lt 1.25</m>. 
         So
         <md>
-          <mrow>0\amp \lt  x\lt 2</mrow>
-          <mrow>0\amp \lt  x^2\lt 4\amp\amp \text{(Squared each term.)}</mrow>
-          <intertext>Since <m>0\lt x\lt 2</m>, we can add <m>0</m>,
-            <m>x</m> and <m>2</m>, respectively,
-            to each part of the inequality and maintain the inequality.</intertext>
-          <mrow>0\amp \lt  x^2+x\lt 6</mrow>
-          <mrow>-1\amp \lt  x^2+x-1\lt 5\amp\amp \text{(Subtracted 1 from each part.)}</mrow>
+          <mrow>0\frac34 amp \lt  x\lt \frac54</mrow>
+          <mrow>\frac{9}{16}\amp \lt  x^2\lt \frac{25}{16}\amp\amp \text{(Squared each term.)}</mrow>
+          <intertext>Adding the above two inequalities, we get:</intertext>
+          <mrow>\frac{21}{16}\amp \lt  x^2+x\lt \frac{45}{16}</mrow>
+          <mrow>\frac{5}{16}\amp \lt  x^2+x-1\lt \frac{29}{16}\amp\amp \text{(Subtracted 1 from each part.)}</mrow>
         </md>
       </p>
 
       <p>
         In <xref ref="eq_lim4" text="global">Inequality</xref>,
         we wanted <m>\abs{x-1}\lt \varepsilon/\abs{x^2+x-1}</m>.
-        The above shows that given any <m>x</m> in <m>[0,2]</m>,
+        The above shows that given any <m>x</m> in <m>(0.75,1.25)</m>,
         we know that
         <md>
-          <mrow>x^2+x-1 \amp \lt 5 \amp\amp \text{which implies that}</mrow>
-          <mrow>\frac15 \amp \lt \frac{1}{x^2+x-1} \amp\amp \text{which implies that}</mrow>
-          <mrow number="yes" xml:id="eq_lim5">\frac{\varepsilon}5 \amp \lt \frac{\varepsilon}{x^2+x-1}\amp\amp</mrow>
+          <mrow>\abs{x^2+x-1} \amp \lt \frac{29}{16} \amp\amp \text{which implies that}</mrow>
+          <mrow>\frac{16}{25} \amp \lt \frac{1}{\abs{x^2+x-1}} \amp\amp \text{which implies that}</mrow>
+          <mrow number="yes" xml:id="eq_lim5">\frac{16\varepsilon}{25} \amp \lt \frac{\varepsilon}{\abs{x^2+x-1}}\amp\amp</mrow>
         </md>.
       </p>
 
       <p>
-        So we set <m>\delta \lt \epsilon/5</m>.
+        So we set <m>\delta \lt 16\epsilon/25</m>, on the interval <m>(0.75, 1.25)</m>.
         This ends our scratch-work, and we begin the formal proof
         (which also helps us understand why this was a good choice of <m>\delta</m>).
       </p>
 
       <p>
-        Given <m>\varepsilon</m>, let <m>\delta \lt \varepsilon/5</m>.
+        Given <m>\varepsilon\gt 0</m>, let <m>\delta \lt \min\{1/4, 16\varepsilon/25\}</m>.
         We want to show that when <m>\abs{x-1}\lt \delta</m>,
         then <m>\abs{(x^3-2x)-(-1)}\lt \varepsilon</m>.
         We start with <m>\abs{x-1}\lt \delta</m>:
         <md>
           <mrow>\abs{x-1} \amp \lt  \delta</mrow>
-          <mrow>\abs{x-1} \amp \lt  \frac{\varepsilon}{5}</mrow>
+          <mrow>\abs{x-1} \amp \lt  \frac{16\varepsilon}{25}</mrow>
           <mrow>\abs{x-1} \amp \lt  \frac{\varepsilon}{\abs{x^2+x-1}} \amp\amp  \text{(}<xref ref="eq_lim5" text="global">Inequality</xref>, x\text{ near 1)}</mrow>
           <mrow>\abs{x-1}\cdot \abs{x^2+x-1} \amp \lt  \varepsilon</mrow>
           <mrow>\abs{x^3-2x+1} \amp \lt  \varepsilon</mrow>
           <mrow>\abs{(x^3-2x)-(-1)} \amp \lt \varepsilon</mrow>
         </md>,
-        which is what we wanted to show.
+        which is what we wanted to show. 
+        (Note that the condition <q><m>x</m> near <m>1</m></q> is satisfied by the fact that <m>\delta\lt 1/4</m>,
+        so that <m>3/4\lt x\lt 5/4</m>.)
         Thus <m>\lim_{x\to 1}(x^3-2x) = -1</m>.
       </p>
     </solution>

ptx/sec_sequences.ptx

@@ -991,6 +991,12 @@
               <m>\lim\limits_{n\to\infty} c\cdot a_n = c\cdot L</m>
             </p>
           </li>
+
+          <li>
+            <p>
+              If <m>g</m> is continuous at <m>L</m>, then <m>\lim\limits_{n\to\infty}g(a_n)=g(L)</m>.
+            </p>
+          </li>
         </ol>
       </p>
     </statement>