Science:Math Exam Resources/Courses/MATH100/December 2011/Question 01 (d)

MATH100 December 2011

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q1 (h)  •  Q1 (i)  •  Q1 (j)  •  Q1 (k)  •  Q1 (l)  •  Q1 (m)  •  Q1 (n)  •  Q2 (a)  •  Q2 (b)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q5 (d)  •  Q5 (e)  •  Q6  •  Q7  •  Q8  •  

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Question 01 (d)

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. Suppose f and g are continuous functions such that g(3) = 2 and $\displaystyle \lim _{x\to 3}(xf(x)+g(x))=1.$ Find f(3).

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 hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
What limit laws can you use to separate the limit?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Our goal is to find f(3). Since f(x) is continuous we know that $f(3)=\lim _{x\to 3}f(x)$ and thus finding f(3) amounts to finding the limit. Using the limit laws we have $1=\lim _{x\to 3}(xf(x)+g(x))=\lim _{x\to 3}x\lim _{x\to 3}f(x)+\lim _{x\to 3}g(x).$ Using that both functions x and g(x) are continuous, we have that $\lim _{x\to 3}x=3$ and $\lim _{x\to 3}g(x)=g(3)=2.$ Plugin this into the above equality, we find $1=3\lim _{x\to 3}f(x)+2$ and hence $\lim _{x\to 3}f(x)=-{\frac {1}{3}}.$ By the first remark above we have $\displaystyle f(3)=-{\frac {1}{3}}.$

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