Science:Math Exam Resources/Courses/MATH101/April 2012/Question 01 (i)/Solution 1

Recall the Maclaurin (power) series of sin(x) is

$\sin(x)=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\dots$

Note that the first two terms cancel exactly with the polynomial in the numerator. Thus,

${\begin{aligned}\lim _{x\rightarrow 0}{\frac {\sin(x)-x+x^{3}/6}{\sin(x^{5})}}&=\lim _{x\rightarrow 0}{\frac {x^{5}/(5!)-x^{7}/(7!)+\cdots }{\sin(x^{5})}}\\&=\lim _{x\to 0}\left({\frac {1}{5!}}{\frac {x^{5}}{\sin(x^{5})}}-{\frac {x^{2}}{7!}}{\frac {x^{5}}{\sin(x^{5})}}+\dots \right)\end{aligned}}$

Now, as shown in Hint 2,

$\lim _{x\to 0}{\frac {y}{\sin(y)}}=1$

we can also say that

$\lim _{x\rightarrow 0}{\frac {x^{5}}{\sin(x^{5})}}=1$

So the first term of the limit becomes ${\frac {1}{5!}}$. The remaining terms all have an $x^{5}/\sin(x^{5})$ term which evaluates to $1$, but also have an extra power of $x$ which will make the entire term go to 0. So the solution to the limit is ${\frac {1}{5!}}$.

This question demonstrates the strength of power series in more difficult calculus problems.