#$&* course Mth 173 10/14 19:50
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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). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* 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? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Problem 1.7.24 submitted through text_11 Problem 1.7.20 submitted through text_11 It is told that the value of the function f(x) = sin(x)/x at x = 0 is 1/2. For a function to be continuous at a particular x value, the value of the function at that x should be equal to the limit of the function tending to that value. That is for f(x) to be continuous f(0) = lim x tending to 0 f(x). Now to calculate the value of the limit of the function as x tends to 0, we perform the following steps We find the value of the function for x values in the vicinity of x = 0, that is for x = 0.001, x = 0.00001 for right hand limit and x = -0.001 and x = -0.0001 for left hand limit. The value of the limit will be the value to which the value of the function is tending too. We see that the function tends to 1 as x tends to 0 from either side. Another way is using the L hospital rule, that is since the function has a 0/0 inequality for x = 0, thus we may calculate the value of the limit by differentiating the numerator and denominator separately and then calculating the report. Thus lim x tending to 0 g(x)/h(x) = lim x tending to 0 g’(x)/h’(x). Thus lim x tending to 0 sin(x)/x = lim x tending to 0 cos(x)/1 = 1 Thus we see that the limit of the f(x) as x tends to 0 is 1 and the value of the function at 0 is ½ thus the function is not continuous. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query problem Find lim (cos h - 1 ) / h, h -> 0. What is the limit and how did you get it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: To calculate lim h tending to 0 ( cos h - 1 ) / h. One way to calculate is by finding the values of the function using a calculator for all x values within the vicinity of h = 0, both on the left hand as well as the right hand side and calculating the value of the function to which the function for the x values in the vicinity tend to. In doing so we calculating the value of the limit to be 0. Another way is, cos(h) = 1 - 2 * sin^2(h/2), thus cos h - 1 = -2 sin^2(h/2). Thus the function reduces to -2 sin^2(h/2) / h To calculate the limit we multiply divide the function by h / 4. Thus the function will also be -2[sin^2(h/2) / ( h^2 / 4)] * (h/4). The limit of the function is lim h tending to 0 (sin^2(h/2)/(h^2/4)) * lim h tending to 0 (h/4) Thus the value of the limit of the function is 1 * 0 = 0. Another way is using the L hospital rule, that is since the function has a 0/0 inequality for x = 0, thus we may calculate the value of the limit by differentiating the numerator and denominator separately and then calculating the report. Thus lim x tending to 0 g(x)/h(x) = lim x tending to 0 g’(x)/h’(x). Thus lim x tending to ( cos h - 1 ) / h = lim x tending to 0 - sin(h) / 1 = 0. confidence rating #$&*: 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. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `q Query Add comments on any surprises or insights you experienced as a result of this assignment. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK 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. "" " 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. "" " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!