Science:Math Exam Resources/Courses/MATH101/April 2017/Question 10/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: $\color {blue}-{\frac {1}{12}}$