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