Science:Math Exam Resources/Courses/MATH100/December 2012/Question 03 (b)

MATH100 December 2012

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Question 03 (b)
Short-Answer Questions. Questions 1-4 are short-answer questions. Put your answers in the boxes provided. Simplify your answers as much as possible, and show your work. Each question is worth 3 marks, but not all questions are of equal difficulty. If $4x-9\leq f(x)\leq x^{2}-4x+7$ for $x\geq 0$, find $\lim _{x\to 4}f(x)$

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
This question looks ready for an application of the squeeze theorem which states if functions f,g,h are such that $g(x)\leq f(x)\leq h(x)$ and if $\lim _{x\to a}g(x)=L=\lim _{x\to a}h(x)$, then $\lim _{x\to a}f(x)=L$.

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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. Following the advice of the hint since $\lim _{x\to 4}(4x-9)=4(4)-9=16-9=7$ and $\lim _{x\to 4}(x^{2}-4x+7)=(4)^{2}-4(4)+7=16-16+7=7$ As these two limits are the same and our function $f(x)$ is bounded below and above by the two functions above, we have that $\lim _{x\to 4}f(x)=7$ by the squeeze theorem.

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