Science:Math Exam Resources/Courses/MATH101 A/April 2025/Question 05/Solution 1

Recall the following Taylor series: $$e^{t}=\sum _{n=0}^{\infty }{\frac {t^{n}}{n!}},\ \sin(t)=\sum _{n=0}^{\infty }{\frac {(-1)^{n}t^{2n+1}}{(2n+1)!}}.$$ Then $${\begin{aligned}e^{x^{4}}=\sum _{n=0}^{\infty }{\frac {x^{4n}}{n!}}=1+x^{4}+{\frac {x^{8}}{2}}+{\frac {x^{12}}{3!}}+...\\\sin(x^{2})=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+2}}{(2n+1)!}}=x^{2}-{\frac {x^{6}}{3!}}+{\frac {x^{10}}{5!}}+...\\x^{2}\sin(x^{2})=\sum _{n=0}^{\infty }{\frac {(-1)^{n}x^{4n+4}}{(2n+1)!}}=x^{4}-{\frac {x^{8}}{3!}}+{\frac {x^{12}}{5!}}+...\end{aligned}}$$ so $$e^{x^{4}}-x^{2}\sin(x^{2})-1=({\frac {x^{8}}{2}}+{\frac {x^{12}}{6}}+...)-(-{\frac {x^{8}}{6}}+{\frac {x^{12}}{5!}}+...)={\frac {2x^{8}}{3}}+{\frac {5x^{12}}{5!}}+....$$ Hence, we have that $${\frac {e^{x^{4}}-x^{2}\sin(x^{2})-1}{x^{8}}}={\frac {2}{3}}+{\frac {5x^{4}}{5!}}+...$$ and so taking the limit as $x\rightarrow 0$ gives ${\frac {2}{3}}$. We could also compute the limit using summation notation directly. $$e^{x^{4}}-x^{2}\sin(x^{2})-1=\sum _{n=1}^{\infty }{\frac {x^{4n}}{n!}}-\sum _{m=1}^{\infty }{\frac {(-1)^{m-1}x^{4m}}{(2m-1)!}}=\sum _{n=1}^{\infty }{\frac {x^{4n}}{n!}}+\sum _{m=1}^{\infty }{\frac {(-1)^{m}x^{4m}}{(2m-1)!}}=\sum _{n=2}^{\infty }{\Big (}{\frac {1}{n!}}+{\frac {(-1)^{n}}{(2n-1)!}}{\Big )}x^{4n},$$ so we get $$\lim _{x\rightarrow \infty }{\frac {e^{x^{4}}-x^{2}\sin(x^{2})-1}{x^{8}}}=\lim _{x\rightarrow \infty }\sum _{n=2}^{\infty }{\Big (}{\frac {1}{n!}}+{\frac {(-1)^{n}}{(2n-1)!}}{\Big )}x^{4n-8}={\frac {1}{2}}+{\frac {1}{6}}={\frac {2}{3}}.$$