Science:Math Exam Resources/Courses/MATH 180/December 2017/Question 11 (b)

MATH 180 December 2017

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Question 11 (b)
The function $$P(x)={\frac {1}{4}}xe^{-x/2},\quad x\geq 0,$$ is an example of a Gamma distribution, a function that may be used to describe certain probabilities. (b) Find the horizontal asymptote of ${\textstyle P(x)}$.

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 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
We need to find the limit as ${\textstyle x}$ tends to positive infinity.

Hint 2
Use L'Hôpital's rule.

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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. By the Hint, we need to find $$\lim _{x\to \infty }P(x)=\lim _{x\to \infty }{\dfrac {x}{4e^{x/2}}}.$$ Note that we have written ${\textstyle P(x)}$ as a quotient. Since $$\lim _{x\to \infty }x=\infty$$ and $$\lim _{x\to \infty }4e^{x/2}=\infty ,$$ the limit is in the indeterminate form of ${\textstyle {\dfrac {0}{0}}}$, and we need to use L’Hôpital’s rule, as follows: $${\begin{aligned}\lim _{x\to \infty }P(x)&=\lim _{x\to \infty }{\dfrac {x}{4e^{x/2}}}\\&=\lim _{x\to \infty }{\dfrac {(x)'}{(4e^{x/2})'}}\\&=\lim _{x\to \infty }{\dfrac {1}{4e^{x/2}\cdot {\frac {1}{2}}}}\\&=\lim _{x\to \infty }{\dfrac {1}{2e^{x/2}}}\\&=0.\end{aligned}}$$ Notice that in the third line, we use the chain rule with ${\textstyle f(x)=e^{x}}$, ${\textstyle g(x)={\frac {x}{2}}}$, ${\textstyle f'(x)=e^{x}}$ and ${\textstyle g'(x)={\frac {1}{2}}}$ to obtain $$(4e^{x/2})'=4(e^{x/2})'=4(f(g(x)))'=4f'(g(x))g'(x)=4e^{x/2}\cdot {\frac {1}{2}}.$$ Therefore, the horizontal asymptote is ${\textstyle y=0}$. Answer: the horizontal asymptote is ${\textstyle {\color {blue}y=0}}$.