Improved problem 3

Exams/Rutgers-Analysis/midterm.cecilia.tex

@@ -1,4 +1,5 @@
 \documentclass{article}
+\usepackage{amsmath}
 \usepackage{amssymb}
 \usepackage{nth}
 \usepackage[utf8]{inputenc}
@@ -64,22 +65,36 @@ \section*{Problem 2}
 \end{eqnarray}
 
 \section*{Problem 3}
-Let $ \delta = \min (1, \frac{\epsilon}{8}) $, note that $ 0 < \delta < 1 $, so $ 0 < \delta^2 < \delta $.
+Let $ \delta = \min (\sqrt{\frac{\epsilon}{2}}, \frac{\epsilon}{8}) $.
 
-Whenever $ | x - 1 | < \delta $, we have
+On one hand, we have:
 
-\begin{eqnarray*}
-    -\delta &< x - 1 &< \delta \\
-    2 -\delta &< x + 1 &< 2 + \delta \\
-    (2 -\delta)^2 &< (x + 1)^2 &< (2 + \delta)^2 \\
-    4 - 4\delta + \delta^2 &< x^2 + 2x + 1 &< 4 + 4\delta + \delta^2 \\
-    -4\delta + \delta^2 &< (x^2 + 2x + 1) - 4 &< 4\delta + \delta^2 \\
-    -4\delta &< (x^2 + 2x + 1) - 4 &< 5\delta \\
-    -\frac{\epsilon}{2} &< (x^2 + 2x + 1) - 4 &< \frac{5\epsilon}{8} \\
-    -\epsilon &< (x^2 + 2x + 1) - 4 &< \epsilon
-\end{eqnarray*}
+\begin{align*}
+  |x - 1| &< \delta \\
+  |x - 1| &< \sqrt{\frac{\epsilon}{2}} \\
+  |(x-1)^2| &< \frac{\epsilon}{2} &\text{square is okay because LHS $\ge$ 0} \\
+\end{align*}
+
+On the other hand, we have:
+
+\begin{align*}
+  |x - 1| &< \delta \\
+  |x - 1| &< \frac{\epsilon}{8} \\
+  |4(x-1)| &< \frac{\epsilon}{2} \\
+\end{align*}
+
+Therefore, by the triangle inequality, we have:
+
+\begin{align*}
+   & |x^2 + 2x + 1 - 4 | \\
+  =& |x^2 - 2x + 1 + 4x - 4| \\
+  =& |(x-1)^2 + 4(x-1)| \\
+  <& |(x-1)^2| + |4(x-1)| \\
+  <& \frac{\epsilon}{2} + \frac{\epsilon}{2} \\
+  =& \epsilon
+\end{align*}
 
-So we conclude that $ \lim_{x \to 1} x^2 + 2x + 1 = 4 $.
+Therefore we proved that $ \lim\limits_{x \to 1} x^2 + 2x + 1 = 4 $ by the definition of the limit, it is possible to find a $ \delta $ such that $ |x - 1| < \delta $ implies $ |x^2 + 2x + 1 - 4| < \epsilon $.
 
 \section*{Problem 4}