Science:Math Exam Resources/Courses/MATH102/December 2011/Question 02 (a) i

MATH102 December 2011

  •  Q1 (a) i  •  Q1 (a) ii  •  Q1 (b) i  •  Q1 (b) ii  •  Q1 (b) iii  •  Q1 (c)  •  Q2 (a) i  •  Q2 (a) ii  •  Q2 (a) iii  •  Q2 (b) i  •  Q2 (b) ii  •  Q2 (b) iii  •  Q2 (c)  •  Q3  •  Q4 (i)  •  Q4 (ii)  •  Q4 (iii)  •  Q4 (iv)  •  Q4 (v)  •  Q4 (vi)  •  Q5  •  Q6  •  Q7 (i)  •  Q7 (ii)  •  Q7 (iii)  •  Q8 (i)  •  Q8 (ii)  •  Q8 (iii)  •  Q9  •  

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Question 02 (a) i
Short-Answer Problems. A correct answer in the box gives full marks. For partial marks work needs to be shown. Note: If a limit does not exist, indicate whether it approaches $+\infty$ or $-\infty$. Find the limit $\displaystyle A=\lim _{x\to \infty }{\frac {2x^{4}-3x^{3}+5}{2x^{3}-5x^{4}}}$

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 is the dominating term in the numerator and the denominator?

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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. $\displaystyle A=\lim _{x\to \infty }{\frac {2x^{4}-3x^{3}+5}{2x^{3}-5x^{4}}}$ To evaluate this limit, first we divide both the numerator and denominator by the largest power of $x$ that appears in this fraction. In this case, $x^{4}$ is the largest power. $\displaystyle A=\lim _{x\to \infty }{\frac {2-3/x+5/x^{4}}{2/x-5}}.$ Now all that's left to finish the problem, is calculating the limits separately: ${\begin{aligned}\displaystyle A&=\lim _{x\to \infty }{\frac {2-3/x+5/x^{4}}{2/x-5}}\\&={\frac {\lim _{x\to \infty }2-3/x+5/x^{4}}{\lim _{x\to \infty }2/x-5}}&={\frac {2}{-5}}\end{aligned}}$ Therefore, $\displaystyle {\color {blue}A=-{\frac {2}{5}}}$.

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