Science:Math Exam Resources/Courses/MATH101/April 2018/Question 09 (i)/Solution 1

Using the Maclaurin series for the exponential function and for the cosine function, we have

$$2e^{-2x^{2}}=2\left(1+(-2x^{2})+{\frac {(-2x^{2})^{2}}{2!}}-{\frac {(-2x^{2})^{3}}{3!}}+\cdots \right)$$

and

$$\cos(x^{2})=1-{\frac {(x^{2})^{2}}{2!}}+{\frac {(x^{2})^{4}}{4!}}+\cdots .$$

Thus we have

$${\begin{aligned}\lim _{x\to 0}{\frac {2-4x^{2}-2e^{-2x^{2}}}{1-\cos(x^{2})}}&=\lim _{x\to 0}{\frac {-{\frac {2(-2x^{2})^{2}}{2!}}+{\frac {2(-2x^{2})^{3}}{3!}}+\cdots }{{\frac {(x^{2})^{2}}{2!}}-{\frac {(x^{2})^{4}}{4!}}+\cdots }}=\lim _{x\to 0}{\frac {-{\frac {8x^{4}}{2!}}+{\frac {16x^{6}}{3!}}+\cdots }{{\frac {x^{4}}{2!}}-{\frac {x^{8}}{4!}}+\cdots }}\\&=\lim _{x\to 0}{\frac {-{\frac {8}{2!}}+{\frac {16x^{2}}{3!}}+\cdots }{{\frac {1}{2!}}-{\frac {x^{4}}{4!}}+\cdots }}={\frac {-{\frac {8}{2!}}}{\frac {1}{2!}}}=-8.\end{aligned}}$$

Answer: The correct answer is ${\textstyle {\color {blue}-8}}$.