Query 11

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course MTH 173

5/31/2013 at 11:47PM

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:

Since x^2 is always positive, it is impossible for the denominator to be 0. Since it is not divisible by zero, the function is defined for every value of x. The function has a smooth graph and is continuous.

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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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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 continuous. This is because the limiting value of the function at x = 0 is 1 and not ½.

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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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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:

h = .1

(cos(h) - 1) / h = -0.05

h = .01

(cos(h) - 1) / h = -0.005

h = .001

(cos(h) - 1) / = -0.0005

The limit is zero. This can be found by finding the y-value on the y-axis.

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3

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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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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 understand the assignment. Everything was well-covered in the notes.

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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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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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