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

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