Math 1500: Tips for Final Exam Part 2
Published: Thu, 04/05/12
If you missed the previous tips I sent out, you can see them at this link:
I just got this email from a student in Dr. Gupta's class (Anything in boldface is my interpretation of what that means):
Student's email: "I just finished my last class with prof. Gupta she more or less said
everything from midterm on will be on the exam as well as some limits
and such but what caught my attn. was she mentioned the squeeze theorm
in her review. I tried to ask her if it was on the exam and all she said
was i should know it to calculate limits. being the first time ive
heard her mention it i have a good feeling its on the exam."
Study my two examples of Squeeze Theorem in Lesson 2 (question 17).
Student's email: "Also from the req. text book p.190 she pointed out examples and stressed that we should know like the lim X -> 0 Sin(x)/X =1 or P194 Sin7X/4X = 7/4"
Student's email: "Also from the req. text book p.190 she pointed out examples and stressed that we should know like the lim X -> 0 Sin(x)/X =1 or P194 Sin7X/4X = 7/4"
Study how to do Trig Limits as I show in Lesson 2 (question 16).
Student's email: "Another point was on p106 lim X-> 0 X^2 Sin(1/X) DNE and told
us to read up why? this is also where the squeeze therom was mentioned."
The student got the info garbled here. That limit does exist and equals 0 and can be proved using the Squeeze Theorem. What does not exist is the limit as x approaches 0 of sin(1/x) because sin(+/- infinity) does not approach any particular value (it merely cycles between -1 and 1 forever). However, the limit as x approaches 0 of x^2 sin(1/x) does exist because:
-1 <= sin(1/x) <= 1
Therefore:
-x^2 <= x^2 sin(1/x) <= x^2
Since limit as x approaches 0 of -x^2 and x^2 are both 0, the limit as x approaches 0 of x^2 sin(1/x) = 0 also, by Squeeze Theorem.
Student's email: "She mentioned curve sketching for max/min word problems. Also on a side note that confused me a little she mentioned somthing
about being given the critial points and using the second derivative
test to find the equation, because the CP=0 . im not sure if i miss
heard but it didnt make any sense to me."
This may mean they intend to give you a question where you are specifically told to use the Second Derivative Test to prove that a Critical Point is either a local max or local min. See question 20 in my Practise Problems for Lesson 9 (Curve-Sketching) for a discussion of the Second Derivative Test and how to use it. It is also possible that she was just referring to the need to prove the Critical Point is a maximum or minimum in curve-sketching and max/min word problems which I prefer to do by the Sign Diagram for the First Derivative. However, do study the Second Derivative Test to be safe.