#$&*
course Mth 173
6/16 10:06 pm
011. `query 11
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Question: `q problem 1.7.6 (was 1.11.4) continuity of x / (x^2+2) on (-2,2)is the function continuous on the given interval and if so, why?
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Your solution:
Yes, because the denominator is not zero from any x value from -2 to 2
confidence rating #$&*:0
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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:
The function is not defined at x=0 because that would be the denominator 0
And when x = 0 the function is closer to 1 instead of 1/2
confidence rating #$&*:2
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Given Solution:
** Division by zero is not defined, so sin(x) / x cannot exist at x = 0. The function is, however, not defined at x = 0 by sin(x) / x; the definition says that at x = 0, the function is equal to 1/2.
It remains to see what happens to sin(x) / x as x approaches zero. Does the function approach its defined value 1/2, in which case the value of the function at x = 0 would equal its limiting value x = 0 and the function would be continuous; does it approach some other number, in which case the limiting value and the function value at x = 0 would not the equal and the function would not be continuous; or does the limit at x = 0 perhaps not exist, in which case we could not have continuity.
Substituting small nonzero values of x into sin(x) / x will yield results close to 1, and the closer x gets to 0 the closer the result gets to 1. So we expect that the limiting value of the function at x = 0 is 1, not 1/2. It follows that the function is not continuous. **
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Self-critique (if necessary):
OK
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Self-critique Rating: OK
@&
When x = 0 the value of the function is 1/2.
When x is near 0 the value is near 1.
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Question: `q Query problem
Find lim (cos h - 1 ) / h, h -> 0.
What is the limit and how did you get it?
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Your solution:
Table of h values:
H:
1 fn= 1
2 fn= .270
3 fn= -.139
4 fn = -.247
the limit would be x goes to -infinity with h values greater than 0
confidence rating #$&*:0
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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):
????????? I used increasing values that were greater than 0 and farther away from 0, why did you use numbers closer to 0? Is that what the h -> 0 means?
I’m also not sure why since you go - and positive numbers close to 0 why the limit is 0?
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The values you used are 1, 2, 3, 4. These values aren't close to zero.
As you should understand if you understand the cosine function, the cosine is periodic with period 2 pi.
1, 2, 3 and 4 are each a significant portion of 2 pi. These values cannot be considered close to zero.
On the other hand, .1, .01, .001 are each a small portion of 2 pi, and are appropriate to thinking about this limit.
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Self-critique Rating:0
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The values of h are all positive.
However you could equally well use negative values of h which are close to 0.
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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):
I found that sin(x) / x, x<>0; 1/2 for x = 0 means that the expression = ½ at x= 0
I now understand what this means
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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.
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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.
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Self-critique (if necessary):
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Self-critique rating:
#*&!
&#This looks good. See my notes. Let me know if you have any questions. &#