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,

$\begin{array}{rl}\underset{x\to 0}{\lim }\frac{\sin (x)-x+x^3/6}{\sin (x^5)}& =\underset{x\to 0}{\lim }\frac{x^5/(5!)-x^7/(7!)+\cdots }{\sin (x^5)}\\ & =\underset{x\to 0}{\lim }(\frac{1}{5!}\frac{x^5}{\sin (x^5)}-\frac{x^2}{7!}\frac{x^5}{\sin (x^5)}+\dots )\end{array}$

Now, as shown in Hint 2,

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

we can also say that

$\underset{x\to 0}{\lim }\frac{x^5}{\sin (x^5)}=1$

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.