From c8d9561a40bc8cf7c36cbf0456e32e5744aec7fe Mon Sep 17 00:00:00 2001
From: Camil Staps
Date: Tue, 22 Sep 2015 18:10:53 +0200
Subject: Finish assignment 3

---
 assignment3.tex | 12 +++++++++---
 1 file changed, 9 insertions(+), 3 deletions(-)

(limited to 'assignment3.tex')

diff --git a/assignment3.tex b/assignment3.tex
index cb6165a..be6e3e8 100644
--- a/assignment3.tex
+++ b/assignment3.tex
@@ -41,8 +41,9 @@
                 Finally, we know $x=\sin\frac{\sqrt3}2$ gives $x=(2k+\frac13)\cdot\pi$ or $x=(2k+\frac23)\cdot\pi$ with $k\in\mathbb Z$. The only possible $x$ would then be $(2k+\frac13)\pi$ with $k=0$, so $\arcsin\frac{\sqrt3}2 = \frac\pi3$.
             \item \begin{align*}
                     f'(x) &= \left(\arcsin\frac{2x}{1-x}\right)' \\
-                          &= \frac1{1-\left(\frac{2x}{1-x}\right)^2} \cdot \left(\frac{2x}{1-x}\right)' \\
-                    %todo
+                          &= \frac1{\sqrt{1-\left(\frac{2x}{1-x}\right)^2}} \cdot \left(\frac{2x}{1-x}\right)' \\
+                          &= \frac1{\sqrt{1-\left(\frac{2x}{1-x}\right)^2}} \cdot \frac{(1-x)\cdot2 - 2x\cdot-1}{(1-x)^2} \\
+                          &= \frac2{(1-x)^2\sqrt{1-\left(\frac{2x}{1-x}\right)^2}}.
                 \end{align*}
         \end{enumerate}
 
@@ -138,7 +139,12 @@
 
                 The graph of $f$ intersect the $y$ axis at $x=0$; this gives the point $(0,f(0)) = (0, 2)$.
                 
-            \item %todo
+            \item \begin{align*}
+                    \lim\limits_{x\to\infty} \frac{(x-2)^2}{x+2} &= \lim\limits_{x\to\infty} \frac{1-\frac4x+\frac4{x^2}}{\frac1x+\frac2{x^2}} = \infty.\\
+                    \lim\limits_{x\to-\infty} \frac{(x-2)^2}{x+2} &= \lim\limits_{x\to-\infty} \frac{1-\frac4x+\frac4{x^2}}{\frac1x+\frac2{x^2}} = -\infty.
+                  \end{align*}
+
+                  There exists no two-sided limit for $x\to-2$. This can be seen from the graph of $f$: on the right of $x=-2$, $f$ goes to $\infty$, while on the left, $f$ goes to $-\infty$.
                 
             \item \begin{align*}
                     f(x)    &= \frac{(x-2)^2}{x+2} = \frac{x^2 - 4x + 4}{x+2}.\\
-- 
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