Tuesday, December 8, 2009

I've Reached My Limit

Types Problems I Did Not Understand
1) #9 from pg 95 lim (xcscx+1)/(xcscx) and other problems like this involving the dividing of trig functions
x->0

2) #5 on pg 95

3) # 14 on pg 92 f(x)= |x-2| at x=1

3 comments:

  1. for #9 on pg 95, you have to seperate the fraction into (xcscx/xcscx)+(1/xcscx).
    xcscx/xcscx is equal to 1, and 1/xcscx is also equal to sinx/x (1/cscx is equal to sinx).1 + the limit as x approaches 0 of sinx is 1... 1+1=2.

    for #5,make the top have a common denominator of 2(2+x). Forst solve the top and you get -x/4+2x, then divide it by the x that was on the bottom of the fraction (you really have to multiply by 1/x instead). then you get -1/4+2x. when you plug in 0 for x (since the original equation said to find the limit as x approaches 0), you get -1/4.

    And i cant help you with number 14 since i did not understand that one myself..

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  2. 3) Graphically, it's the |x| shifted over the right 2. This means at x=1, the part of the graph you are seeing is the y=-x+2 line (as opposed to the y=x-2 line). Does that help you finish?

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  3. Yes Exactly Hwang,
    For question 3 here is what you can assume:
    |x-2|, is the same as the lines x-2, and 2-x (-x+2) above the x-axis (positive y values). This means that at x=1 there should be a slope of -1, (hence, -x+2).

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