Convergent values of series containing variables approache zero

Question

Example 7 in Thomas' Calculus Early Transcendentals 14th edition p.614 demonstrates how to use Taylor series to find a limit involving an indeterminate form:

$$ \lim_{x\to 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right). $$

After applying Taylor expansion to $\sin(x)$ and using some algebraic manipulation, the author came up with the following equation:

$$ \lim_{x\to 0}\left(\frac{1}{\sin(x)} - \frac{1}{x}\right) = \lim_{x\to 0}x\frac{\frac{1}{3!} - \frac{x^2}{5!} + \frac{x^4}{7!} - ...}{1 - \frac{x^2}{3!} + \frac{x^4}{5!} -...} $$

After that he concluded that the limit is $0$. But I'm not able to figure out why it's true because as $x$ approaches zero, how do we know the fraction on the right side of equation above approaches some constant and not infinity since both numerator and denominator are series that contain a variable (not a constant) approaching zero?

Answer 1

One way to define the sine function is to let $\sin(x) = \sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!}$. The theory of power series can then be used to show that $\sin(x)$ is defined on all of $\mathbb R$ (in fact all of $\mathbb C$) and is an infinitely differentiable function satisfying the usual properties.

To compute $\lim_{x \to 0} \left(\frac{1}{\sin(x)}-\frac{1}{x}\right)$ using this definition note that $$ \begin{split} \frac{1}{\sin(x)}-\frac{1}{x} = \frac{x-\sin(x)}{x\cdot\sin(x)} &= \frac{\sum_{n=1}^\infty(-1)^{n+1}x^{2n+1}/(2n+1)!}{\sum_{n=0}^\infty (-1)^n x^{2n+2}/(2n+1)!}\\ \end{split} $$ Now letting $m=n-1$ we see that $$ \sum_{n=1}^\infty(-1)^{n+1}x^{2n+1}/(2n+1)! = x^3f_1(x), \quad f_1(x) = \sum_{m=0}^\infty \frac{(-1)^mx^{2m}}{(2m+3)!}. $$ We also have $$ \sum_{n=0}^\infty (-1)^n x^{2n+2}/(2n+1)!=x^2f_2(x), \quad f_2(x) = \sum_{n=0}^{\infty}\frac{(-1)^n\cdot x^{2n}}{(2n+1)!} $$ so that $$ \frac{1}{\sin(x)} - \frac{1}{x} = \frac{x^3f_1(x)}{x^2f_2(x)} = x\cdot\frac{f_1(x)}{f_2(x)}. $$ But $f_1(x)$ and $f_2(x)$ are functions defined by power series which converge on all of $\mathbb R$, so they are infinitely differentiable functions. In particular they are continuous, and so $\lim_{x\to 0} f_1(x)/f_2(x) = f_1(0)/f_2(0)=\frac{1/3!}{1!} = 1/6$, and hence $$ \lim_{x\to 0}\left( \frac{1}{\sin(x)} - \frac{1}{x} \right) = \lim_{x \to 0} x\cdot\frac{f_1(x)}{f_2(x)} = \lim_{x\to 0} x\cdot\lim_{x\to 0}\frac{f_1(x)}{f_2(x)} =0\cdot\frac{1}{6} = 0 $$

Answer 2

We shall try to see this a bit differently.

It is very easy to see that $$\lim_{x \rightarrow +\infty}\frac{1}{x}=0.$$ Suppose that $$\lim_{x \rightarrow +\infty}\bigg( \frac{1}{\sin(x)}-\frac{1}{x} \bigg)$$ exists.

Then, $\lim_{x \rightarrow +\infty}\frac{1}{x}=0$ along with the existence of the given limit will guarantee that $$\lim_{x \rightarrow +\infty}\frac{1}{\sin(x)}=\lim_{x \rightarrow +\infty} \csc(x)$$ exists, which is absurd. This can be understood very clearly once you visualize that the functions $\csc(x)$ and $\big( \frac{1}{\sin(x)}-\frac{1}{x} \big)$ behave similarly as you go far away from $x=0.$ Indeed, from the below graph, it is very clear.

The red coloured curve is that of the function $\big( \frac{1}{\sin(x)}-\frac{1}{x} \big),$ while blue coloured is that of the function $\csc(x).$ As $x$ moves away from the origin, we have the following scene:

Hence, the given limit does not exist!

Answer 3

After the edit, the limit is: $$\lim_{x\to 0}\left(\frac{1}{\sin x}-\frac1x\right) $$ As suggested by the author of the book, the best way to solve the limit is using the Taylor expansion of the sine function: $\sin x =x-x^3/6+o(x^3)$. You have: $$ \frac{1}{\sin x}-\frac1x=\frac{x-\sin x}{x\sin x}\sim\frac{x^3}{6x^2}=\frac{x}{6} \to 0$$