Week+03+Sept+1-5

September 3rd 2008
Hey, guys! I haven't remembered my password yet but here is my contribution to the 5th period textbook from Wednesday, the 3rd! And, I know the order is a liiiittle crazy, but I've been clicking and dragging for a very long time to no avail. Sorry! But, all examples are random AND numbered, so it shouldn't make a big difference :) Good luck you guys.

September 4th 2008
We began class by going over several homework problems,some of which may appear on this page at a later date! The aim for the day was to figure out two trigonometric limits both of which are fundamental to all trig. limits. These limits are: The first limit is derived algebraically and the second algebraically and the methods are shown in the slides below.

Tomorrow we will learn how to use these basic limits to evaluate other, more complex, trigonometric limits.

September 5th 2008
Today, as promised by yesterday's scribe, we worked on a trig. limits worksheet which is accessible using the link below. The solutions to these problems will appear shortly.

=Trig Limits Homework Problems= -Number 4. Bryan Haywood

Lim secx-1/ xsecx = Lim ((1/cosx-1)-1)/ x(1/cosx) = x-›0 x-›0 Lim ((1-cosx)/cosx)/ (x/cosx) = Lim ((1-cosx)/cosx) * (cosx/x)= x-›0 x-›0 Lim (1-cosx)/x = 0 x-›0
 * 23. by ali ross**


 * Number 2 by Gabi Smith**



Lim· 3x/sin5x= x->0 Lim 3* x/sin5x = x->0 Lim 3* 5x/5sin5x= x->0 Lim 3* 1/5sin5x/5x= x->0 Lim 3* 1/5= 3/5 x->0
 * Number 6 - Jay Strickland**
 * Number 18 - Brittany Nehman**
 * Problem 3 - Curran Sidhu **


 * Number 4 by Emma Dekock**

okay..here it goes. brilliant #4. it's calling this worksheet problem 15 for some reason. my life.

Lim sin 3x/sin 2x x —› 0

=

Lim 1/2 * sin 3x/x x —› 0

=

Lim 1/2 * 3sin 3x/3x x —› 0

=

Lim 1/2 * 3 x —› 0

= 3/2 aka la respuesta correcta aka "what it is"


 * Problem 7-> Siobhan Donnelly** **Problem #12 -- Travis Dunlap**


 * Problem 23- Grant Moody**


 * Problem #9 -- Rachel Neill**

Lim sin² 3x/ 5x² x —> 0

Lim sin 3x/ 5x × sin 3x/ x x —> 0

Lim sin 3x/ 5x × Lim 3/5 × Lim sin 3x/ x × Lim 3 x —> 0 x —> 0 x —> 0 x —> 0

1 × 3/5 × 1 × 3

= 9/5


 * Drew Kerr- Problem 8**


 * 25) By Andrew Bezek**
 * Problem 1 by Andrew Spitz**
 * 11) Por Karen Wilmer**

a) Lim 2x x-> 0 tan3x

= (Lim stays the same for each step) __cos3x (2x)__ (multiply this by 3/3) sin3x

= __(6x) cos3x__ (divide this by 2/2) 3sin3x

= __(3x) cos3x__ (3/2) sin3x

= __1__ __(3/2) sin3x__ (3x) cos3x

= __1__ __(3/2)__ 1

= __2__ 3
 * 13 - Kathleen Cornett[[image:trig_#_13.GIF width="411" height="247" align="left"]]**


 * Problem 21 Emily Hixon**




 * Problem 25-Alexandra Rossetti**
 * Ethan Lucas - Pr**[[image:Cal_Trig_Limit_WS.JPG width="458" height="397" align="left"]]**oblem #6**