assignment 11

date 23/8/10" "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: It is bc when plugged into the function it makes a continious line through this point.

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 I see kinda of the range in which you are speaking of.

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Self-critique rating #$&*3

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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 solution is that when the values given are submitted into the equation we have alot of values that are close to 1 but never reach it so this means that the limit is 1. Also this means that the function is not continious because it has values on the pos and neg side and the limit means that it not continious

confidence rating #$&* 2.5

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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):Still a lil fuzzy.

The point is that as x approaches zero, sin(x) / x approaches 1, as you have indicated; however the value of the function at x = 0 is 1/2, so the value of the function is not equal to its limiting value.

The function just goes along, nicely approaching a limiting value of 1, but then its value suddenly, at one point only, drops to 1/2. Not continuous.

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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 is 1 we get 1 when h is 2 we get .5 when h is 3 we get .33 so we find that the limit of this is 0 because we can go the other way and find that the limit if falling toward 0 closer and closer but does not cross it.

When h = 2 we get (cos(2) - 1) / 2. cos(2) is about .8, so the expression becomes (.8 - 1) / 2 = -.1, not .5.

Similarly your values for h = 1 and h = 3 are not correct.

It isn't clear how you got values 1, .5 and .33.

confidence rating #$&* 2.5

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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):should have used smaller numbers as indicated. (easier graphing)

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Self-critique rating #$&*2.5

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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):

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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."

See my notes.

Also be sure you understand that my note in the Gradebook indicated that you need to do qa's 11 - 16, not queries 11-16. Nothing wrong with doing the queries, but with deadlines coming up you need to focus first on those qa's.

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