Calculus Problem Solving Drill 07: Trigonometric Limits and Continuity No. of 0 Instruction: () Read the problem statement and answer choices carefully. () Do your work on a separate sheet of paper. (3) Choose the best answer. (4) Go back to the tutorial to review the core concept tutorial, if necessary.. Which of the following functions is not continuous everywhere? 4 A. f( ) 4 7 + 3.6 B. f( ) sin 4 C. f( ) + D. f ( ) E. They are all continuous everywhere Polynomial functions are continuous everywhere. Look for features like square roots or denominators that could be made 0. This is a basic variation on sin, which is continuous everywhere. Look for features like square roots or denominators that could be made 0. C. Correct! This rational function has a denominator that can be made to equal zero, which means that there will be a vertical asymptote a discontinuity. Eponential functions are continuous everywhere. Look for features like square roots or denominators that could be made 0. One of the functions is discontinuous Most often, a function will have a discontinuity when it has a feature that can lead to undefined values. The most common of these are square roots and denominators that are variable epressions that can be made 0. C is the choice with one of these features when, its denominator is zero. Furthermore, we can recognize the other three choices as functions that are continuous everywhere, and thereby rule them out: A is a polynomial function. B is a basic variant on sin. D is an eponential function.
No. of 0 Instruction: () Read the problem statement and answer choices carefully () Work the problems on paper as needed (3). Which of the following functions is continuous everywhere? A. f( ) B. f( ) + + C. f( ) + + 5 D. f( ) E. They are all continuous everywhere If, this function has a discontinuity. Look for a function whose denominator cannot be made 0 with any real values of. B. Correct! The denominator of this function can never be made 0, since it has no real roots. Therefore, this function has no discontinuities! If, this function has a discontinuity. Look for a function whose denominator cannot be made 0 with any real values of. If 0, this function has a discontinuity. Look for a function whose denominator cannot be made 0 with any real values of. Only one of the functions is continuous everywhere. All of these functions look fairly similar they are all rational functions (one polynomial divided by another), and have similar degrees in the numerator and the denominator. To answer the question, we need to remember what causes discontinuities. If a function doesn t have any discontinuities, it must be continuous everywhere. We recall that one of the most common causes of discontinuities is the denominator of the function being 0. All of these functions have a denominator with a variable, so we just have to recognize one of them as having a denominator epression that, no matter what value of you put in, will never equal 0. Let s look at the epressions in each denominator: will be 0 when. + cannot be 0 for any real number! This is because the square of any real number is non-negative, and when you add the result will always be > 0. + + can be factored to ( + ), which is 0 when. can clearly be 0, when 0. The answer is B.
No. 3 of 0 Instruction: () Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 3. Evaluate the following limit. lim cosθ π θ 3 A. 3 B. π C. 3 D. π E. The limit does not eist. A. Correct! cos is a continuous trig function, so you can find the limit just by plugging in the value that the independent variable approaches. cos is a continuous trig function, so you can find the limit just by plugging in the value that the independent variable approaches. Make sure you plugged in this value correctly. cos is a continuous trig function, so you can find the limit just by plugging in the value that the independent variable approaches. Remember that the answer should be the cosine of an angle, not the angle itself. cos is a continuous trig function, so you can find the limit just by plugging in the value that the independent variable approaches. Remember that the answer should be the cosine of an angle, not the angle itself. cos is a continuous trig function, and the limits of continuous functions always eist. You should be able to find the limit just by plugging in the value that the independent variable approaches. We are trying to find the limit of the cosine function, which is continuous everywhere. As a result, we can find the limit by simply plugging in the value the independent variable approaches. π lim cosθ cos 3 π θ 3 When finding the values of trig functions, it helps some students to convert radian angle measures to degrees first. Here, 3 π radians 60. It also might help to draw a 30-60 -90 triangle, with all of its proportions: Hypotenuse Opposite Leg 3 60 Using this diagram, and recalling cos 60. Adjacent leg adj leg that cosθ, we see that hypotenuse
No. 4 of 0 Instruction: () Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 4. Evaluate the following limit. ( cos ) lim sin π A. B. 0.47 C. D. 0 E. The limit does not eist. This function is a simple variant of the sine and cosine functions, and, as a result, it is continuous. You should be able to find the limit simply by plugging in the value the independent variable is approaching. Remember that these epressions mean take the sine and cosine first, and then square each result. C. Correct! This function is a simple variant of the sine and cosine functions, and, as a result, it is continuous. You were able to find the limit simply by plugging in the value the independent variable is approaching. This function is a simple variant of the sine and cosine functions, and, as a result, it is continuous. You should be able to find the limit simply by plugging in the value the independent variable is approaching. This function is a simple variant of the sine and cosine functions, and, as a result, it is continuous. The limits of continuous functions always eist. This function is a simple variant of the sine and cosine functions, and, as a result, it is continuous. You should be able to find the limit simply by plugging in the value the independent variable is approaching. ( ) ( π ) ( π ) lim sin cos sin cos π Before continuing, we recall that an operation like cos actually means take the cosine of the value first and then square the result. This notation can be confusing, and it shows up often in higher math. ( π) ( π) ( ) ( ) sin cos 0 0 The limit equals. Another, slightly quicker way to evaluate this epression is to recall the cosine double-angle formula from trigonometry: cos sin cos θ θ θ Here we have the opposite of this formula: ( ) ( ) ( ) ( ) sin π cos π cos π sin π cosπ
No. 5 of 0 Instruction: () Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 5. Evaluate the following limit. limcsc 0 A. 0 B. C. D. E. The limit does not eist csc is a trig function that is a quotient of two other trig functions. The denominator function is sin, so when sin is 0, csc will have a vertical asymptote. csc is a trig function that is a quotient of two other trig functions. The denominator function is sin, so when sin is 0, csc will have a vertical asymptote. Once you have determined that csc has a vertical asymptote at 0, test values slightly less and greater than 0. If they are very different, than the limit may not eist. csc is a trig function that is a quotient of two other trig functions. The denominator function is sin, so when sin is 0, csc will have a vertical asymptote. E. Correct! csc has a vertical asymptote at 0, and testing values slightly less and greater than 0 reveals that the limit does not eist. csc is a discontinuous function with periodically repeating vertical asymptotes. This is because it can be rewritten as csc sin. Any function with a denominator that can be made 0 is discontinuous. Finding a limit at a vertical asymptote is a different process than elsewhere, so we want to know if the function has a vertical asymptote at 0, the value being approached. To determine this, we see if 0 makes the denominator, sin, equal 0. sin( 0) 0 There is a vertical asymptote where we are taking our limit. To find the limit at a vertical asymptote, plug in numbers slightly less than and more than the value approaches. ( ) ( ) csc 0.0 00.0 csc 0.0 00.0 (Remember that the csc operator is not an option in many calculators. To compute it, input /sin(0.0) into the calculator.) The results are very far apart from one another, so we conclude that the limit does not eist.
No. 6 of 0 Instruction: () Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 6. Evaluate the following limit. ( + ) sin cos θ θ 0 lim θ θ A. 0 B. C. D. E. The limit does not eist Try to find a removable discontinuity (a factor whose denominator approaches 0, but that approaches altogether), and cancel it out. Then plug the value the independent variable approaches back into the simplified function. Try to find a removable discontinuity (a factor whose denominator approaches 0, but that approaches altogether), and cancel it out. Then plug the value the independent variable approaches back into the simplified function. C. Correct! Once you cancel out the removable discontinuity and plug 0 back into the simplified function, you get a value of. Try to find a removable discontinuity (a factor whose denominator approaches 0, but that approaches altogether), and cancel it out. Then plug the value the independent variable approaches back into the simplified function. Try to find a removable discontinuity (a factor whose denominator approaches 0, but that approaches altogether), and cancel it out. Then plug the value the independent variable approaches back into the simplified function. The first thing we should observe about this function is that it has the important removable discontinuity sinθ, which approaches as θ 0. Since that is the value θ approaches in our limit, we should be θ able to cancel out our removable discontinuity as a factor of will not change the value of the answer. sinθ ( cosθ + ) ( θ + ) 0 θ 0 lim lim cos θ θ Once we have canceled out the removable discontinuity, we are free to assume that the result is continuous, and try to find the limit by simply plugging in. lim cos + cos 0 + + θ ( θ ) ( ) 0 Because plugging in gave us a real number, the simplified function was in fact continuous, and the limit equals.
No. 7 of 0 Instruction: () Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 7. Evaluate the following limit. lim cot 0 A. 0 B. C. D. E. The limit does not eist If you rewrite cot as a quotient of two functions, you may be able to locate a removable discontinuity. Cancel it out and then try plugging 0 back into the function. B. Correct! Rewriting cot as a quotient of two functions reveals a removable discontinuity. Once you cancel it out and plug 0 back in for, you get. Good work! If you rewrite cot as a quotient of two functions, you may be able to locate a removable discontinuity. Cancel it out and then try plugging 0 back into the function. If you rewrite cot as a quotient of two functions, you may be able to locate a removable discontinuity. Cancel it out and then try plugging 0 back into the function. If you rewrite cot as a quotient of two functions, you may be able to locate a removable discontinuity. Cancel it out and then try plugging 0 back into the function. We begin by rewriting function. cot as a quotient, which will help us pinpoint the discontinuities in the cos lim cot lim 0 0 sin We don t automatically see the fundamental removable discontinuity, sin. But, we can sort of spot its sin reciprocal,. To find we do a little bit more rearranging, and recall that by switching a sin fraction into the denominator it turns into its reciprocal. cos cos lim lim ( cos ) lim 0 0 0 sin sin sin sin sin Now we have found two eamples of our removable discontinuity. Canceling them out, we have the limit of a continuous function, which we can evaluate by plugging in. limcos cos (0) 0 The limit equals.
No. 8 of 0 Instruction: () Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 8. Consider the following function. f( ) sin At which of the following -values does the function have a discontinuity? I. 0 II. III. π A. I only B. II only C. I and III D. II and III E. I, II, and III The function will have discontinuities whenever the denominator equals 0. Set the denominator equal to zero and solve. Since this is an equation with a periodic function (sin), it will probably have numerous solutions. The function will have discontinuities whenever the denominator equals 0. Set the denominator equal to zero and solve. Since this is an equation with a periodic function (sin), it will probably have numerous solutions. C. Correct! The function will have discontinuities whenever the denominator equals 0, which includes all the angles whose terminal position is on the -ais. The function will have discontinuities whenever the denominator equals 0. Remember that sin is 0 at all the angles whose terminal position is on the -ais. The function will have discontinuities whenever the denominator equals 0. Set the denominator equal to zero and solve. Since this is an equation with a periodic function (sin), it will probably have numerous solutions. The primary causes of discontinuities in functions defined by a single equation are zeroes in the denominator and negatives within a radical. Since there are no roots, we restrict our search to zeroes of the denominator. f( ) sin So, we need to figure out when sin 0. Since sin is a periodic function, this equation will have infinitely many solutions. What are they? The sin of an angle is 0 if the angle s standard position is on the -ais. π, 3π, 5π,... 0, π, 4π,... As demonstrated in the diagram, this happens when the angle is a multiple of π. Of the options we are given, 0 and π fit this criterion. They are the values that make sin 0, therefore they are the places where the function is discontinuous.
No. 9 of 0 Instruction: () Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 9. Consider the following piecewise function. f( ) e < for for lim f( )? A. B. C. e D.- E. The limit does not eist A. Correct! The independent variable is approaching, which is the transition value between the two equations of the piecewise function. We find the limit of both equations as approaches. Since both limits are, this is the answer! The independent variable is approaching, which is the transition value between the two equations of the piecewise function. Find the limit as approaches in the two different equations. The independent variable is approaching, which is the transition value between the two equations of the piecewise function. Find the limit as approaches in the two different equations. The independent variable is approaching, which is the transition value between the two equations of the piecewise function. Find the limit as approaches in the two different equations. The limits of the two equations as approaches are equal, so the limit does eist. The first observation is that the value is approaching is, which is the transition value for the piecewise function it s the -value where the function switches from e to. Since both of these epressions represent functions that are continuous everywhere, we can find the limit as by just plugging into the two epressions. If the outcomes are equal, this is the value of our function. If not, then the limit does not eist. ( ) 0 e e ( ) The values agree, so the limit equals.
No. 0 of 0 Instruction: () Read the problem statement and answer choices carefully () Work the problems on paper as needed (3) 0. Consider the following piecewise function. ( ) k + for 0 f( ) cos + k for > 0 For what value of k will the function be continuous everywhere? A. B. 0 C. D. E. There is no such value of k. at 0 agree. at 0 agree. at 0 agree. at 0 agree. E. Correct! at 0 agree. The resulting equation has no solution, so there is no value of k that will make this function continuous everywhere. at 0 agree. If the functions have the same value at 0, they will connect everywhere. We start by setting 0 in both equations. ( 0 ) ( ) k + k cos 0 + k + k Then, we set these two values equal to make sure that the two sides of the function meet. k k + 0 In trying to solve for k, we realize that there is no solution. This means that no value of k could make the function continuous at 0.