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course Mth 173
2/28 8:17PM
011. `query 11
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Question: `q problem 1.7.6 (was 1.11.4) continuity of x / (x^2+2) on (-2,2)is the function continuous on the given interval and if so, why?
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Your solution:
there is never a division by zero. I believe this is continuous
confidence rating #$&*: 2
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Given Solution:
** The denominator would never be 0, since x^2 must always be positive. So you could never have division by zero, and the function is therefore defined for every value of x. The function also has a smooth graph on this interval and is therefore continuous.
The same is true of the correct Problem 4, which is 1 / `sqrt(2x-5) on [3,4]. On this interval 2x-5 ranges continuously from 2*3-5=1 to 2*4-5=3, so the denominator ranges continuously from 1 to `sqrt(3) and the function itself ranges continuously from 1 / 1 to 1 / `sqrt(3). **
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Self-critique (if necessary): OK
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Self-critique Rating:OK
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Question: `q query problem 1.7.24 5th; 1.7.20 4th (was 1.11.9) continuity of sin(x) / x, x<>0; 1/2 for x = 0. Where is the function continuous and where is it not continuous?
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Your solution:
i dont know what "" x<>0; 1/2 for x = 0. "" means.
????well you can not divide by zero, so it wont be continous at x=0, however i remember the graph of sin and cos went up and down with a period and altitude, so I graphed it, and on my calculator, it looks continous, it just goes up really high at x=0, compared to the rest of the graph
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x <> 0 means x is greater than or less than zero. This would cover all possible values of x except zero. So another way to read it would the 'x not equal to 0'.
Your calculator cannot display a removable discontinuity at a single point. This discontinuity is removable; the limit of sin(x) / x as x -> 0 is 1, so if we define the function to be equal to 1 instead of 1/2 at x = 0, the value of the function will be equal to its limit and the function ibecomes continuous.
Incidentally you should be graphing this function without the use of a calculator. Unless otherwise specified, calculator graphs are inadmissible in solutions (though they can be very useful to check your solutions).
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You should know why the limiting value of sin(x) is 1. It is not so because the calculator says it is so. It is so because of the circular definition of the sine function, and the argument is mainly geometric.
To avoid confusion, I'll use theta instead of x, so I can refer to the unit-circle point corresponding to angle theta as (x, y). It would be very confusing to use x for the argument of the sine, and for the horizontal coordinate of the unit-circle point.
So we'll talk about the limit of sin(theta) / theta, as theta approaches zero.
sin(theta) is the y coordinate of the unit-circle point corresponding to angle theta.
theta is the angle of the line that intersects the unit circle at the point (x, y), the line making angle theta with the positive x axis, theta being measured in radians.
Since theta is measured in radians and the unit circle has radius 1, theta is also the length of the arc of the unit circle from the positive x axis to the intersection point (x, y).
If theta is small, the arc of the unit circle is very nearly a straight line, and is very nearly perpendicular to the x axis. The length of this arc is therefore very nearly to the y coordinate of the point (x, y) of intersection. That y coordinate is sin(theta), so theta is very nearly equal to sin(theta) and sin(theta) / theta is very nearly equal to 1. The closer theta gets to zero, the closer the arc will be to a straight line, and the closer sin(theta) / theta will be to 1. You can make sin(theta) / theta as close to 1 as you wish by making theta small enough. Thus the limit of sin(theta) / theta, as theta approaches zero, is 1.
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See also Class Notes on your DVD's.
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confidence rating #$&*: 1
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Given Solution:
** Division by zero is not defined, so sin(x) / x cannot exist at x = 0. The function is, however, not defined at x = 0 by sin(x) / x; the definition says that at x = 0, the function is equal to 1/2.
It remains to see what happens to sin(x) / x as x approaches zero. Does the function approach its defined value 1/2, in which case the value of the function at x = 0 would equal its limiting value x = 0 and the function would be continuous; does it approach some other number, in which case the limiting value and the function value at x = 0 would not the equal and the function would not be continuous; or does the limit at x = 0 perhaps not exist, in which case we could not have continuity.
Substituting small nonzero values of x into sin(x) / x will yield results close to 1, and the closer x gets to 0 the closer the result gets to 1. So we expect that the limiting value of the function at x = 0 is 1, not 1/2. It follows that the function is not continuous. **
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Self-critique (if necessary):
The graph of this one looks continous when I did, so I am getting confused, maybe I graphed it wrong. So is it continous except at x=0? which it maybe would equal 1/2?
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Self-critique Rating: 2
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Question: `q Query problem
Find lim (cos h - 1 ) / h, h -> 0.
What is the limit and how did you get it?
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Your solution:
when h= .001,.01,.1
-.000000499999958333334722222, -0.0000499996, -0.00499583
it said as h approaches 0, so plugged in numbers approaching 0. the number approaches 0 from the negative side.
confidence rating #$&*: 2
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Given Solution:
** For h = .1, .01, .001 the values of (cos(h)-1 ) / h are -0.04995834722, -0.004999958472, -0.0005. The limit is zero. **
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Self-critique (if necessary): OK
@&
There is also an argument proving this limit, based on the circular definition.
For now, though, the proof for sin(x) / x is enough.
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Self-critique Rating: OK
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Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment.
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Self-critique (if necessary):
It is making me reconsider the meaning of a limit, I always viewed it as 0 being important, but now I see it more as everything approaching 0, and 0 isnt the only important value. and it seems the sin and cos have some interesting graphs/limits.
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That is the case, mostly for sin(x) / x and (cos(x) - 1) / x as x -> 0.
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Self-critique Rating:
STUDENT QUESTION:
Is the limit also where the function becomes discontinuous?
INSTRUCTOR RESPONSE:
A function is continuous at a certain x value if, as you approach that x value, the limiting value of the function is equal to its value at the point.
This is equivalent to the following two conditions:
If the limiting value of a function y = f(x), as you approach a certain x value, doesn't equal the value of the function, then the function is not continuous.
If the function doesn't have a limit at a certain x value, then the function is not continuous at that x value."
Self-critique (if necessary):
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Self-critique rating:
STUDENT QUESTION:
Is the limit also where the function becomes discontinuous?
INSTRUCTOR RESPONSE:
A function is continuous at a certain x value if, as you approach that x value, the limiting value of the function is equal to its value at the point.
This is equivalent to the following two conditions:
If the limiting value of a function y = f(x), as you approach a certain x value, doesn't equal the value of the function, then the function is not continuous.
If the function doesn't have a limit at a certain x value, then the function is not continuous at that x value."
Self-critique (if necessary):
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Self-critique rating:
#*&!
&#Good responses. See my notes and let me know if you have questions. &#