course Mth 173 [zwlַHĸassignment #011
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12:23:46 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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RESPONSE --> There is never going to be a denominator of zero because the x^2 is always positive. You can't divide by zero.
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12:23:49 ** 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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12:27:46 query problem 1.7.16 (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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RESPONSE --> If x = 0 then it is continuous but with the number getting closer to 0 then answer gets closer to 1 which means that it nevers passes through 0 which it is not continuous.
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12:27:48 ** 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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12:27:52 Query Add comments on any surprises or insights you experienced as a result of this assignment.
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