Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (a)
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Question 01 (a) |
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Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Full marks will be given for correct answers placed in the box, but at most 1 mark will be given for incorrect answers. Unless otherwise stated, it is not necessary to simplify your answers in this question. Evaluate or determine that the limit does not exist. |
Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you? |
If you are stuck, check the hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint. |
Hint 1 |
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What happens if you simply plug in -2 in the function? What does it mean? What can you do about that? |
Hint 2 |
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Evaluating at -2 gives 0/0 which is undefined. This means we just don't know anything about what the limit might be (it could be anything). So we need to deal with that. Notice that the function is a rational function (a fraction with polynomials in both numerator and denominator), since each are 0 at t=-2, it means that -2 is a root of each and hence, they can be factorized. |
Checking a solution serves two purposes: helping you if, after having used all the hints, you still are stuck on the problem; or if you have solved the problem and would like to check your work.
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Solution |
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Found a typo? Is this solution unclear? Let us know here.
Please rate my easiness! It's quick and helps everyone guide their studies. When we evaluate the function at we obtain 0/0. Since this is a rational function, it means that we can factor in each of the denominator and numerator: and Hence |