If your solution to stated problem does not match the given solution, you should self-critique per instructions at
http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm
.
Your solution, attempt at solution.
If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.
011. `query 11
Question:
`q problem 1.7.4; 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?
Your solution:
Confidence Assessment:
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). **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q query problem
1.7.32 6th;
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?
Your solution:
Confidence Assessment:
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. **
Self-critique (if necessary):
Self-critique Rating:
Question: `q Query problem
Find lim (cos h - 1 ) / h, h -> 0.
What is the limit and how did you get it?
Your solution:
Confidence Assessment:
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. **
Self-critique (if necessary):
Self-critique Rating:
Question:
`q Query Add comments
on any surprises or insights you experienced as a result of this assignment.
Self-critique (if necessary):
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.</h3>