#$&* course Mth 173 7/20 2:11am 021. `query 21
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Given Solution: `a** The question was about the motion of the particle. The graph of f(t) has at every point a slope of magnitude 1, except at the points where the slope changes. The graph of g(t) has at every point a slope of magnitude 1, except at the points where the slope changes. At points where the slope changes, it changes 'instantly', so at those points the slope is not defined. More formally at these points the quantity (f(t + `dt) - f(t) ) / `dt, whose limit defines the integral, has a different limit when `dt approaches zero through positive values than when it approaches 0 through negative values, so the limit is not defined. At all times t, except where the slope changes sign, | f ' (t) | = 1. Sometimes f ' (x) is positive, sometimes negative. At all times t, except where the slope changes sign, | g ' (t) | = 4. Sometimes g ' (x) is positive, sometimes negative. Thus at all times t, except where the slope changes sign, the speed of the particle is v = sqrt( 1^2 + 4^2 ) = sqrt(17), about 4.1. At clock time t = 0 the x coordinate is f(0) = 2 and the y coordinate is g(0) = 0. At clock time t = 1/2 the x coordinate is f(1/2) = 1.5 and the y coordinate is g(1/2) = 1. So between t = 0 and t = 1/2, the particle moves from (2, 0) to (1.5, 1). At clock time t = 1 the x coordinate is f(1) = 1 and the y coordinate is g( 1 ) = 0 . So between t = 1/2 and t = 1, the particle moves from (1.5, 1) to (1, 0). At clock time t = 1.5 the x coordinate is f(1.5) = .5 and the y coordinate is g(1.5) = -1. So between t = 1 and t = 1.5, the particle moves from (1,0) to (.5,-1). At clock time t = 2 the x coordinate is f(2) = 0 and the y coordinate is g(2) = 0. So between t = 1.5 and t = 2, the particle moves from (.5, -1) to (0, 0). At clock time t = 2.5 the x coordinate is f(2.5) = .5 and the y coordinate is g(2.5) = 1. So between t = 2 and t = 2.5, the particle moves from (0, 0) to (.5, 1). During the next few half-second intervals the particle moves to (1, 0), (1.5, -1) and (2, 0) The first graph is of position vs. time, i.e., x vs. t for a particle moving on the x axis. The slope of the x position vs. time graph is the x component of the velocity of the particle. The first graph has slope -1 for negative values of t. So up to t = 0 the particle is moving to the left at velocity 1. Then when the particle reaches x = 0, which occurs at t = 0 the slope becomes +1, indicating that the velocity of the particle instantaneously changes from -1 to +1 and the particle moves back off to the right. During the 4-second interval the x position therefore changes from x = 2 to x = 0 then back to x = 2. On the second graph the velocity of the particle changes abruptly--instantaneously, in fact--when the graph reaches a sharp point, which it does twice between t = 0 and t = 2, and twice more between t = 2 and t = 4. At these points velocity goes from positive to negative or from negative to positive. So between t = 0 and t = 2, the y component of the position goes from 0 to 1 and back to 0, then to -1 and back up to 0. This occurs as x goes from 2 to 0, so the position of the particle zigzags from (2, 0) to (3/2, 1) to (1, 0) to (1/2, -1) to (0, 0). Between t = 2 and t = 4 the x coordinate moves back to the right, and the particle's path zigzags from (0, 0) to (1/2, 1) to (1, 0) to (3/2, -1) and back to the starting point at (2, 0). Since the x velocity is always 4 times the magnitude of the y velocity (except at the 'turning points' where velocity is not defined), the paths are straight lines. Another way of saying this is that the instantaneous speed, at all points where f ' and g ' are defined, is always v = sqrt( f ' ^2+ g ' ^2 ) = sqrt( 4^2 + 1^2) = sqrt(17) = 4.1, approx., as also specified earlier in this solution. The two paths form a shape on the graph that looks like two diamond shapes. The speed of the particle as it goes from point to point is always 4. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I’m not really sure how you calculated the speed here ------------------------------------------------ Self-critique Rating: 0
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Given Solution: The standard parameterization of a unit circle (i.e., a circle of radius 1) is x = cos(t), y = sin(t), 0 <= t < 2 pi. An ellipse is essentially a circle elongated in two directions. To elongate the circle in such a way that its major and minor axes are the x and y axes we can simply multiply the x and y coordinates by the appropriate factors. An ellipse through the given points can therefore be parameterized as x = 5 cos (t), y = 7 sin (t), 0 <= t < 2 pi. To confirm the parameterization, at t = 0, pi/2, pi, 3 pi/2 and 2 pi we have the respective points (x, y): (5 cos(0), 7 sin(0) ) = (5, 0) (5 cos(pi/2), 7 sin(pi/2) ) = (0, 7) (5 cos(pi), 7 sin(pi) ) = (-5, 0) (5 cos(3 pi/2), 7 sin(3 pi/2) ) = (0, -7) (5 cos(pi), 7 sin(pi) ) = (5, 0). ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): That was much easier than I thought. I was definitely over thinking it ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qQuery 4.8.23 (was 3.8.18). x = t^3 - t, y = t^2, t = 2.What is the equation of the tangent line at t = 2 and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The velocity in the x direction is dx/dt t^3 -t = 3t^2- 1 = 11
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Given Solution: `a** Derivatives are dx/dt = 3t^2 - 1, which at t = 2 is dx/dt = 11, and dy/dt = 2t, which at t=2 is 4. We have x = 6 + 11 t, which solved for t gives us t = (x - 6) / 11, and y = 4 + 4 t. Substituting t = (x-6)/11 into y = 4 + 4 t we get y = 4 + 4(x-6)/11 = 4/ll x + 20/11. Note that at t = 2 you get x = 6 so y = 4/11 * 6 + 20/11 = 44/11 = 4. Alternatively: The slope at t = 2 is dy/dx = dy/dt / (dx/dt) = 4 / 11. The equation of the line thru (6, 4) with slope 4/11 is y - 4 = 4/11 ( x - 6), which simplifies to y = 4/11 x + 20/11. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ??So did you use the parameter equations here? ------------------------------------------------ Self-critique Rating: 2
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Given Solution: `a The velocities in the x and y directions are dx / dt and dy / dt. Since x = cos(t^2) we have dx/dt = -2(t) sin (t)^2. Since y = sin(t^2) we have dy/dt = 2(t) cos (t)^2. Speed is the magnitude of the resultant velocity speed = | v | = sqrt(vx^2 + vy^2) so we have speed = {[-2(t) sin (t)^2]^2 + [2(t) cos (t)^2]^2}^1/2. This simplifies to {4t^2 sin^2(t^2) + 4 t^2 cos^2(t^2) } ^(1/2) or (4t^2)^(1/2) { sin^2(t^2) + cos^2(t^2) }^(1/2) or 2 | t | { sin^2(t^2) + cos^2(t^2) }^(1/2). Since sin^2(theta) + cos^2(theta) = 1 for any theta, this is so for theta = t^2 and the expression simplifies to 2 | t |. The speed at clock time t is 2 | t |. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I did not simplify the expression ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qDoes the particle ever come to a stop? If so when? If not why not? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The particle will come to a stop when v =0 This occurs when t = 0 or when sin^2(t^2) + cos^2(t^2) = 0. But sin^2(t^2) + cos^2(t^2) = 1 so this can’t occur. At t = 0: x = cos(0) = 1 y = sin(0) = 0 which gives the point (1, 0). This is where the particle isn’t moving The particle is not moving at t=0 and only then confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The particle isn't moving when v = 0. v = 2 | t | { sin^2(t^2) + cos^2(t^2) }^(1/2) is zero when t = 0 or when sin^2(t^2) + cos^2(t^2) = 0. However sin^2(t^2) + cos^2(t^2) = 1 for all values of t, so this does not occur. t = 0 gives x = cos(0) = 1 and y = sin(0) = 0, so it isn't moving at (1, 0). The particle is at rest at t = 0, and only at t = 0. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 3.9.33 was 3.9.18 (3d edition 3.9.8) (was 4.8.20) square the local linearization of e^x at x=0 to obtain the approximate local linearization of e^(2x) YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The local linearization is the tangent line So the slope of e^x is the derivative of e^x at x=0 y’(0) = x^0 = 1 so point is (0, 1) tangent line equation is then: y = x+1 we then find the tangent line for e^2x y= 2x + 1 when (x+1) is squared: x^2 + 2x + 1 I’m not sure how these two can be made identical? confidence rating #$&*: 1 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The local linearization is the tangent line. The line tangent to y = e^x at x = 0 is the line with slope y ' = e^x evaluated at x = 0, or slope 1. The line passes through (0, e^0) = (0, 1). The local linearization, or the tangent line, is therefore (y-1) = 1 ( x - 0) or y = x + 1. The line tangent to y = e^(2x) is y = 2x + 1. Thus near x = 0, since (e^x)^2 = e^(2x), we might expect to have (x + 1)^2 = 2x + 1. This is not exactly so, because (x + 1)^2 = x^2 + 2x + 1, not just 2x+1. However, near x = 0 we see that x^2 becomes insignificant compared to x (e.g., .001^ 2 = .000001), so for sufficiently small x we see that x^2 + 2x + 1 is as close as we wish to 2x + 1. ** STUDENT COMMENT I’m not sure I understood the question to be answered. INSTRUCTOR RESPONSE You were asked to square the local linearization The local linearization of e^x is 1 + x, which when squared is 1 + 2 x + x^2. The local linearization of e^(x^2) is 1 + 2 x. The squared local linearization of e^x is 1 + 2 x + x^2, the local linearization of e^(x^2) is 1 + 2 x. Near x = 0 the quantity x^2 becomes insignificant, so near x = 0 the two expressions are nearly identical. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): I wasn’t sure how to get the two to be nearly identical, but it does make sense to say that at numbers near x^2 becomes insignificant ------------------------------------------------ Self-critique Rating: 3 ********************************************* Question: `qWhat do you get when you multiply the local linearization of e^x by itself, and in what sense is it consistent with the local linearization of e^(2x)? Which of the two expressions for e^(2x) is more accurate and why? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: The local linearization of e^x is x + 1 (x+1)*(x+1) = x^2 + 2x + 1 The local linearization of e^ (2x) Is 2x + 1 It is consistent when the values of x are close to 0 I’m not sure how to tell which of the expressions is more accurate? confidence rating #$&*: 0 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The local linearization of e^(2x) is y = 2x + 1. The square of the local linearization 1 + x of e^x is y = (x + 1)^2 = x^2 + 2x + 1 . The two functions differ by the x^2 term. Near x = 0 the two graphs are very close, since if x is near 0 the value of x^2 will be very small. As we move away from x = 0 the x^2 term becomes more significant, giving the graph of the latter a slightly upward concavity, which for awhile nicely matches the upward concavity of y = e^(2x). The linear function cannot do this, so the square of the local linearization of e^x more closely fits the e^(2x) curve than does the local linearization of e^(2x). ** STUDENT COMMENT I see what I’ve been missing. The question is which tang line follows the actual curve more accurate and I see why this whould be the y =e^x local linearization squared. Would this not have something to do with adding the factor exponential growth to the local linearization, of a exponential Fn? INSTRUCTOR COMMENT: Good speculation. The exponential function, by definition, grows exponentially. However we cannot calculate the exact value of an exponential function at a given numerical value of x, except at x = 0. The point here is to find approximate values of the exponential function, first by a local linearization, then by adding a quadratic term to help us follow the curvature of the exponential graph. Later (2d semester) we see that we can keep adding high-power terms (e.g., multiples of x^3, then of x^4, etc.) we can approximate the effects of higher and higher derivatives to obtain functions that approximate the exponential better and better. Our approximation functions will be polynomials, which can be evaluated accurately, and we will obtain expressions for just how accurate we can expect the approximations obtained from these polynomials to be. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): Ok this makes sense, ?????is this kind of like an error factor or something like that? ------------------------------------------------ Self-critique Rating: 3
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Given Solution: `a** In this problem you are referred to the equation T = 2 pi sqrt( g / L). and told that g remains constant while L varies. You are asked to find an expression for the resulting change in T. Within this context the given equation gives T as a function of L. The rate of change of T with respect to L is the derivative dT/dL. Roughly speaking, when you multiply dT / dL by a small change in L, you are multiplying the rate of change of T by the change in L, which gives you the approximate change in T. Another way of saying this is dT/dL * `dL = `dT, where `dT is the linear approximation of the change in T. The main thing we need to calculate is dT / dL: To start with, `sqrt(L/g) can be written as `sqrt(L) / `sqrt(g). It's a good idea to make this separation because L is variable, g is not. So dT / dL = 2 `pi / `sqrt(g) * [ d(`sqrt(L) ) / dL ]. [ d(`sqrt(L) ) / dL ] is the derivative of `sqrt(L), or L^(1/2), with respect to L. So [ d(`sqrt(L) ) / dL ] = 1 / 2 L^(-1/2) = 1 / (2 `sqrt(L)). Thus dT / dL = 2 `pi / [ `sqrt(g) * 2 `sqrt(L) ] = `pi / [ `sqrt(g) `sqrt(L) ] . This is the same as T / (2 L), since T / (2 L) = 2 `pi `sqrt(L / g) / (2 L) = `pi / (`sqrt(g) `sqrt(L) ). Now since dT / dL = T / (2 L) we see that the differential is `dT = dT/dL * `dL or `dT = T / (2 L) * `dL. ** STUDENT QUESTION I am having problems with problems that pertain to proving something equals something else were in the notes can I go back and review this concept? INSTRUCTOR RESPONSE Applications occur throughout this text, and there is no one approach to solving application problems. The best results come from a lot of practice, and from feedback you get on your solutions. This problem is an application of the differential. The only way to learn how to apply the differential is by first understanding what the differential means, in general. Then, to be a bit repetitive, I'll say again that understanding comes with extensive practice. I suggest that you give #22 another shot after looking over the solution below. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): This problem is very difficult for me to understand even after re-reading your solution… ------------------------------------------------ Self-critique Rating: 0
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Given Solution: `a** `dT = T / (2 L) * `dL can be rearranged to the form `dT / T = 1/2 `dL / L. If we want `dT / T to be less than .02, then 1/2 `dL / L must be less than .02, so `dL / L must be less than .04. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: OK ********************************************* Question: `qQuery problem 4.7.4 (3d edition 4.8.9) graphs similar to -x^3 and x^3 at a. What is the sign of lim{x->a} [ f(x)/ g(x) ]? How do you know that the limit exists and how do you know that the limit has the sign you say it does? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Since one graph is negative and one is positive, dividing the two would result in -1 So the limit would be -1 near a point? confidence rating #$&*: 2 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** If one graph is the negative of the other, as appears to be the case, then for any x we would have f(x) / g(x) = -1. So the limit would have to be -1. It doesn't matter that at x = 0 we have 0 / 0, because what happens AT the limiting point doesn't matter, only what happens NEAR the limiting point, where 'nearness' is unlimited by always finite (i.e., never 0). ** GOOD STUDENT COMMENT The limiting pt isn’t as important as what is adjacent to the limiting pt. It can’t be 0. INSTRUCTOR RESPONSE Good. To put it even more strongly: The actual point isn't important at all; only the neighborhood of the point (the region adjacent to the point, as you put it nicely) is relevant to the limit. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): ok ------------------------------------------------ Self-critique Rating: Ok ********************************************* Question: `qQuery 4.7.8 (3d edition 3.10.8) lim{x -> 0} [ x / (sin x)^(1/3) ].What is the given limit and how did you obtain it? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: I used L’Hopital’s rule f(x) = x f’(x) = 1 g(x) = (sin x)^1/3 g’(x) = 1/3 cos (x) * sin(x)^-2/3 f ' (x) / g ' (x) = 1 / (1/3 cos(x) * sin(x)^(-2/3) ) = 3 sin(x)^(2/3) / cos(x). as x --> 0 sin(x) --> 0 the limiting value of f ' (x) / g ' (x) is 0. It follows from l'Hopital's Rule that the limit of f(x) / g(x) is also zero confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a As x -> 0 both numerator f(x) = x and the denominator g(x) = sin(x)^(1/3) both approach 0 as a limit. So we use l'Hopital's Rule f ' (x) = 1 and g ' (x) = 1/3 cos(x) * sin(x)^(-2/3), so f ' (x) / g ' (x) = 1 / (1/3 cos(x) * sin(x)^(-2/3) ) = 3 sin(x)^(2/3) / cos(x). Since as x -> 0 we have sin(x) -> 0 the limiting value of f ' (x) / g ' (x) is 0. It follows from l'Hopital's Rule that the limiting value of f(x) / g(x) is also zero. STUDENT QUESTION I am having trouble with defining the limiting value of the problem. I want to say it is less than zero but the answer says it is zero. Can you explain were I might have this wrong?????
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Given Solution: `a** The local linearization of the numerator is just y = x. The denominator doesn't have a local linearization at 0; rather it approaches infinite slope. The derivative of the denominator sin(x)^(1/3) is cos(x) * 1/3 sin(x^(-2/3) ) = 1/3 * cos(x) / (sin(x))^(2/3). As x approaches zero sin(x) approaches zero, and so does the 2/3 power of sin(x). So the denominator of the derivative approaches zero, so the derivative cannot be evaluated at x = 0. Thus sin(x)^(1/3) does not have a local linearization at x = 0. Its derivative approaches infinity as x approaches 0. So as x -> 0 the ratio of the denominator function to the numerator function increases without bound, making the values of numerator / denominator approach zero. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: Ok ********************************************* Question: `qWhat are the local linearizations of x and sin(x)^(1/3) and how do they allow you to answer the preceding question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Of x: Is simply y = x The denominator actually approaches infinite slope. The derivative of sin(x)^(1/3) = cos(x) * 1/3 sin(x^(-2/3) ) = 1/3 * cos(x) / (sin(x))^(2/3). the derivative approaches zero, so the derivative cannot be evaluated at x = 0. sin(x)^(1/3) does not have a local linearization at x = 0. Its derivative approaches infinity as x approaches 0. Because the ratio increases, the ratio approaches zero confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The local linearization of the numerator is just y = x. The denominator doesn't have a local linearization at 0; rather it approaches infinite slope. The derivative of the denominator sin(x)^(1/3) is cos(x) * 1/3 sin(x^(-2/3) ) = 1/3 * cos(x) / (sin(x))^(2/3). As x approaches zero sin(x) approaches zero, and so does the 2/3 power of sin(x). So the denominator of the derivative approaches zero, so the derivative cannot be evaluated at x = 0. Thus sin(x)^(1/3) does not have a local linearization at x = 0. Its derivative approaches infinity as x approaches 0. So as x -> 0 the ratio of the denominator function to the numerator function increases without bound, making the values of numerator / denominator approach zero. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: Ok #*&! ********************************************* Question: `qWhat are the local linearizations of x and sin(x)^(1/3) and how do they allow you to answer the preceding question? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: Of x: Is simply y = x The denominator actually approaches infinite slope. The derivative of sin(x)^(1/3) = cos(x) * 1/3 sin(x^(-2/3) ) = 1/3 * cos(x) / (sin(x))^(2/3). the derivative approaches zero, so the derivative cannot be evaluated at x = 0. sin(x)^(1/3) does not have a local linearization at x = 0. Its derivative approaches infinity as x approaches 0. Because the ratio increases, the ratio approaches zero confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: `a** The local linearization of the numerator is just y = x. The denominator doesn't have a local linearization at 0; rather it approaches infinite slope. The derivative of the denominator sin(x)^(1/3) is cos(x) * 1/3 sin(x^(-2/3) ) = 1/3 * cos(x) / (sin(x))^(2/3). As x approaches zero sin(x) approaches zero, and so does the 2/3 power of sin(x). So the denominator of the derivative approaches zero, so the derivative cannot be evaluated at x = 0. Thus sin(x)^(1/3) does not have a local linearization at x = 0. Its derivative approaches infinity as x approaches 0. So as x -> 0 the ratio of the denominator function to the numerator function increases without bound, making the values of numerator / denominator approach zero. ** &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary): OK ------------------------------------------------ Self-critique Rating: Ok #*&!#*&!