Week+02

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Monday 8/23
First test of the semester in this class. You will find a copy of the test,along with Ms. Gentry's key, on the Tests page of this wiki.

Tuesday 8/24
Homework last night was exploratory problem set 2.1 from page 33.

The figure below shows the function f(x) =(2x^2 -5x -2) / (x-2)

a) Show that f(2) takes the indeterminate form 0 /0 . Explain why there is no value for f(2). Substituting 2 for x in both the denominator and the numerator gives zero for both parts. f(2) has no value because division by zero is not defined.

b) The number y = 3 is the limit as x approaches 2 . Make a table of values for each 0.001 unit of x from 1.997 to 2.003. Is it true that f(x) stays close to 3 when x is kept close to 2 but not equal to 2?

We can see that if x is kept within .003 units of 2, f(x) remains within .006 units of 3. c) How close to 2 would you have to keep x for f(x) to stay within .0001 units of 3? Within .00001 units of 3? How could you keep f(x) arbitrarily close to 3 just by keeping x close enough to 2, but not equal to 2? If x is kept within .00005 units of 2, f(x) remains within .0001 units of 3. If x is kept within .000005 units of 2, f(x) remains within .00001 units of 3.
 * x || f(x) ||
 * 1.997 || 2.994 ||
 * 1.998 || 2.996 ||
 * 1.999 || 2.998 ||
 * 2 || error ||
 * 2.001 || 3.002 ||
 * 2.002 || 3.004 ||
 * 2.003 || 3.006 ||

If x is kept, very close to 2, (lets call this distance from 2 the letter delta), f(x) will be kept within twice the distance delta from 3.

d) The missing point at x = 2 is called a __removable discontinuity.__ Why do you suppose this name is used? The discontinuity can be removed by assigning f(2) = 3. f(x) =(2x^2 -5x -2) / (x-2) can be reduced to f(x) = 2x-1 if x is not equal to 2, by factoring the numerator and reducing common factors. Substituting 2 into 2x - 1 yields 3, the limit of f(x) as x approaches 2.

Wednesday 8/25
Lots of talk about limits

Thursday 8/26
Today we used the Nspires to confirm refute some properties of limits. The worksheet is shown below.

Friday 8/27
Today we finished up the explanation of the limit of a composition using the functions f(x) = x^2 and g(x) = sinx. We found that (i) the limit of f(x) as x approaches pi/4 is (pi/4) squared or.196349.. (ii) the limit of g(x) as x app roaches pi/4is square root 2 over 2 (iii) the limit of f(g(x)) as x x approaches pi/4 is 1/2. We then looked for the relationship between these numbers and discovered that 1/2 is the square of root 2 over 2. i.e Limit of f(g(x), as x approaches pi/4, is f(limit of g(x) as x approaches pi/4). The slides from the rest of class are shown below.