Science:Math Exam Resources/Courses/MATH101/April 2017/Question 10

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MATH101 April 2017

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Question 10
Evaluate $\lim _{x\to 0}{\frac {\cos x-e^{-x^{2}/2}}{x^{3}\sin x}}$ . A calculator-ready answer is sufficient.

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
Use the Taylor series to evaluate the limit.

Hint 2
(For an alternative solution) Use L'Hôpital's rule.

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 1
Recall the Maclaurin series of $\cos x$ , $\sin x$ and $e^{x}$ ; ${\begin{aligned}\cos x&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{(2n)!}}x^{2n}=1-{\frac {1}{2!}}x^{2}+{\frac {1}{4!}}x^{4}-{\frac {1}{6!}}x^{6}+\cdots ,\\\sin x&=\sum _{n=0}^{\infty }(-1)^{n}{\frac {1}{(2n+1)!}}x^{2n+1}=x-{\frac {1}{3!}}x^{3}+\cdots ,\\e^{x}&=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}}=1+x+{\frac {1}{2!}}x^{2}+{\frac {1}{3!}}x^{3}+\cdots .\end{aligned}}$ Using these series, the numerator and denominator of the given fraction can be written as ${\begin{aligned}\cos x-e^{-x^{2}/2}&=\left(1-{\frac {1}{2!}}x^{2}+{\frac {1}{4!}}x^{4}-{\frac {1}{6!}}x^{6}+\cdots \right)-\left(1+\left(-{\frac {x^{2}}{2}}\right)+{\frac {1}{2!}}\cdot \left(-{\frac {x^{2}}{2}}\right)^{2}+{\frac {1}{3!}}\left(-{\frac {x^{2}}{2}}\right)^{3}+\cdots \right)\\&=\left(1-{\frac {1}{2}}x^{2}+{\frac {1}{24}}x^{4}-{\frac {1}{6!}}x^{6}+\cdots \right)-\left(1-{\frac {x^{2}}{2}}+{\frac {1}{8}}{x^{4}}-{\frac {1}{48}}x^{6}+\cdots \right)\\&=\left({\frac {1}{24}}-{\frac {1}{8}}\right)x^{4}-\left({\frac {1}{6!}}-{\frac {1}{48}}\right)x^{6}+\cdots ,\end{aligned}}$ and $x^{3}\sin x=x^{3}\left(x-{\frac {1}{3!}}x^{3}+\cdots \right)=x^{4}-{\frac {1}{3!}}x^{6}+\cdots .$ Then, the given fraction can be written as ${\begin{aligned}{\frac {\cos x-e^{-x^{2}/2}}{x^{3}\sin x}}&={\frac {\left({\frac {1}{24}}-{\frac {1}{8}}\right)x^{4}-\left({\frac {1}{6!}}-{\frac {1}{48}}\right)x^{6}+\cdots }{x^{4}-{\frac {1}{3!}}x^{6}+\cdots }}\\&={\frac {\left({\frac {1}{24}}-{\frac {1}{8}}\right)-\left({\frac {1}{6!}}-{\frac {1}{48}}\right)x^{2}+\cdots }{1-{\frac {1}{3!}}x^{2}+\cdots }}.\end{aligned}}$ Here, the last equality follows from dividing both the numerator and denominator by $x^{4}$ . Observes that as $x$ goes to $0$ , the numerator converges to $\left({\frac {1}{24}}-{\frac {1}{8}}\right)$ and the denominator converges to $1$ . Therefore, the given limit has the value ${\frac {1}{24}}-{\frac {1}{8}}=-{\frac {1}{12}}$ . Answer: $\color {blue}-{\frac {1}{12}}$

Solution 2
This is an alternative solution. For MATH101 students, the first solution is recommended. Actually, Solution 1 is simpler than Solution2. Using $\lim _{x\to 0}{\frac {\sin x}{x}}=1$ , we have ${\begin{aligned}\lim _{x\to 0}{\frac {\cos x-e^{-x^{2}/2}}{x^{3}\sin x}}=\lim _{x\to 0}{\frac {\cos x-e^{-x^{2}/2}}{x^{4}\cdot {\frac {\sin x}{x}}}}=\lim _{x\to 0}{\frac {\cos x-e^{-x^{2}/2}}{x^{4}}}\cdot {\frac {1}{\lim _{x\to 0}{\frac {\sin x}{x}}}}=\lim _{x\to 0}{\frac {\cos x-e^{-x^{2}/2}}{x^{4}}}\end{aligned}}$ Recall L'Hôpital's rule. Let $f$ and $g$ be differentiable functions. If $\lim _{x\to a}f(x)=\lim _{x\to a}g(x)=0$ or $\pm \infty$ , then we have $\lim _{x\to a}{\frac {f(x)}{g(x)}}=\lim _{x\to a}{\frac {f'(x)}{g'(x)}}$ . We apply this rule several times, to further evaluate the last term above; ${\begin{aligned}\lim _{x\to 0}{\frac {\cos x-e^{-x^{2}/2}}{x^{4}}}&=\lim _{x\to 0}{\frac {(\cos x-e^{-x^{2}/2})'}{(x^{4})'}}=\lim _{x\to 0}{\frac {-\sin x+xe^{-x^{2}/2}}{4x^{3}}}\\&=\lim _{x\to 0}{\frac {(-\sin x+xe^{-x^{2}/2})'}{(4x^{3})'}}=\lim _{x\to 0}{\frac {-\cos x+e^{-x^{2}/2}-x^{2}e^{-x^{2}/2}}{12x^{2}}}\\&=\lim _{x\to 0}{\frac {-\cos x+e^{-x^{2}/2}}{12x^{2}}}-\lim _{x\to 0}{\frac {x^{2}e^{-x^{2}/2}}{12x^{2}}}\\&=\lim _{x\to 0}{\frac {(-\cos x+e^{-x^{2}/2})'}{(12x^{2})'}}-\lim _{x\to 0}{\frac {e^{-x^{2}/2}}{12}}\\&=\lim _{x\to 0}{\frac {\sin x-xe^{-x^{2}/2}}{24x}}-{\frac {1}{12}}={\frac {1}{24}}\left(\lim _{x\to 0}{\frac {\sin x}{x}}-\lim _{x\to 0}e^{-x^{2}/2}\right)-{\frac {1}{12}}=-{\frac {1}{12}}\end{aligned}}$ Answer: $\color {blue}-{\frac {1}{12}}$