Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit?

The limit is
$$\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)$$
which I’m aware can be rearranged to obtain the indeterminate $\dfrac{0}{0}$, but in an attempt to avoid L’Hopital’s rule (just for fun) I tried using the fact that $\sin x\approx x$ near $x=0$. However, the actual limit is $\dfrac{1}{3}$, not $0$.

In this similar limit, the approximation reasoning works out.

Solutions Collecting From Web of "Why doesn't using the approximation $\sin x\approx x$ near $0$ work for computing this limit?"

If we take one more term in the Taylor expansion:

\begin{align}
\sin x&\approx x-\frac{x^3}6+\cdots\\
\sin^2 x&\approx x^2-2x\frac{x^3}6+\cdots\\
\frac 1{\sin^2 x}&\approx\frac 1{x^2-x^4/3}\\
&=\frac 1{x^2}\cdot\frac 1{1-x^2/3}\\
&=\frac 1{x^2}+\frac 1{3-x^2}\\
\lim_{x\to 0}\left[\frac 1{\sin^2 x}-\frac 1{x^2}\right]&=\lim_{x\to 0}\left[\frac 1{x^2}+\frac 1{3-x^2}-\frac 1{x^2}\right]\\
&=\lim_{x\to 0}\frac 1{3-x^2}\\
&=\frac 1 3
\end{align}


To see where the first-order expansion went wrong, it’s necessary to keep track of where the error term goes:

\begin{align}
\sin x&= x+\text{O}(x^3)\\
\sin^2 x&=x^2+2x\text{O}(x^3)+\text{O}(x^3)^2\\
&=x^2+\text{O}(x^4)+\text{O}(x^6)\\
&=x^2+\text{O}(x^4)\\
\frac 1{\sin^2 x}&=\frac 1{x^2+\text{O}(x^4)}\\
&=\frac 1{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\
\frac 1{\sin^2 x}-\frac 1{x^2}&=\frac 1{x^2}\left[\frac 1{1+\text{O}(x^2)}-1\right]\\
&=\frac 1{x^2}\cdot\frac{1-1+\text{O}(x^2)}{1+\text{O}(x^2)}\\
&=\frac{\text{O}(x^2)}{x^2}\cdot\frac 1{1+\text{O}(x^2)}\\
&=\text{O}(1)
\end{align}

Thus the $\sin x\approx x$ approximation is not accurate enough to estimate even the constant term of the expression in the limit. (Note that it does allow us to say that there are no $\text{O}(n^{-1})$ or bigger terms, so the limit probably won’t diverge.)

In this answer it is shown, using only pre-calculus methods, that
$$
\lim_{x\to0}\frac{\sin(x)-x}{x^3}=-\frac16
$$
This is necessary and not derivable from $\lim\limits_{x\to0}\frac{\sin(x)}x=1$.

It then follows that
$$
\begin{align}
\lim_{x\to0}\left(\frac1{\sin^2(x)}-\frac1{x^2}\right)
&=\lim_{x\to0}\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}\\
&=\lim_{x\to0}\frac{x-\sin(x)}{x^3}\lim_{x\to0}\frac{x+\sin(x)}{x}\lim_{x\to0}\frac{x^2}{\sin^2(x)}\\
&=\frac16\cdot2\cdot1\\[3pt]
&=\frac13
\end{align}
$$

I have pointed this out earlier also on MSE (see link1, link2, link3, link4), but it seems that this point needs to be retold and retold till it becomes a tautology like $1 = 1$.

The rigorous meaning of the statement “$\sin x \approx x$ for small $x$” is that $$\lim_{x \to 0}\frac{\sin x}{x} = 1\tag{1}$$ Thus using the approximation $\sin x \approx x$ while evaluating certain limits means using the limit formula $(1)$ in your calculations. There is no more meaning attached to $\sin x \approx x$ other than above limit formula and one should not even try to attach any more meaning to it as far as the evaluation of limits is concerned.

Thus the use of $\sin x \approx x$ in the linked question by OP regarding limit of $(1/x – 1/\sin x)$ is also wrong (although by luck it does produce right answer).

Moreover the accepted answer here (by 2012rcampion) tries to propagate another fallacy which is that the approximation $\sin x \approx x$ is not good enough in the current context and perhaps a better approximation is desired. This sort of gives the message that the approximation $\sin x \approx x$ is somehow good enough for the linked question. Sorry!! this is simply not correct. If we go down this path we face the following question: how do we know which approximation is good enough in a given context? There is no satisfying answer to this.

Something must be mentioned about this “higher order of approximations” based on Taylor series expansions. When we use an approximation like $$f(a + h) \approx f(a) + hf'(a) + \frac{h^{2}}{2!}f”(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a)\tag{2}$$ its rigorous meaning is $$\lim_{h \to 0}\dfrac{f(a + h) – \left\{f(a) + hf'(a) + \dfrac{h^{2}}{2!}f”(a) + \cdots + \dfrac{h^{n – 1}}{(n – 1)!}f^{(n – 1)}(a)\right\}}{h^{n}} = \frac{f^{(n)}(a)}{n!}\tag{3}$$ which also written as $$f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f”(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + o(h^{n})\tag{4}$$ Note that there is no $\approx$ sign in the above equation so that it represents an exact formula and the expression $o(h^{n})$ represents a function $g(h)$ such that $g(h)/h^{n} \to 0$ as $h \to 0$. Thus using Taylor series expansions like $(2)$ while calculating limits means using the equation $(3), (4)$ in an exact manner and not as an approximation. Limit processes are as exact as $2 + 2 = 4$ and not like $\pi \approx 3.1415$.

In case of Taylor series expansion there is another interpretation possible. If $f$ is a function such that it has an expression in the form of a convergent Taylor series around point $a$ then we write $$f(a + h) = f(a) + hf'(a) + \frac{h^{2}}{2!}f”(a) + \cdots + \frac{h^{n}}{n!}f^{(n)}(a) + \cdots\tag{5}$$ Such functions are technically called analytic. This relation is also an exact formula and the RHS is an infinite series in powers of $h$. When using such series for evaluation of limits we don’t need to use the expressions like $o(h^{n})$ but then the theoretical justification for using infinite series in limit processes is the following theorem:

A power series allows the following operations at an interior point of its region of convergence:

  1. Take limits term by term
  2. Term by term integration with respect to variable whose powers are used
  3. Term by term differentiation with respect to variable whose powers are used.

If one is aware of the concept of uniform convergence (and thereby knows the proof of the theorem above regarding power series) then this approach of using infinite power series for limit evaluation is fully rigorous and there is simply no need to think in terms of any approximations or things like $o(h^{n})$ and just use the series (and do algebraic simplification) and take limits.


As far as the answer to this question is concerned the right approach is the one given by robjohn and there is no way to improve upon his answer. However I would like to mention the use of formula $(1)$ in the current context (or in crude language “use the approximation $\sin x \approx x$”). We can write
\begin{align}
L &= \lim_{x \to 0}\frac{1}{\sin^{2}x} – \frac{1}{x^{2}}\notag\\
&= \lim_{x \to 0}\frac{x^{2} – \sin^{2}x}{x^{2}\sin^{2}x}\notag\\
&= \lim_{x \to 0}\frac{x^{2} – \sin^{2}x}{x^{4}}\cdot\left(\frac{x}{\sin x}\right)^{2}\notag\\
&= \lim_{x \to 0}\frac{x^{2} – \sin^{2}x}{x^{4}}\cdot 1^{2}\notag\\
&= \lim_{x \to 0}\frac{x^{2} – \sin^{2}x}{x^{4}}\notag
\end{align}
This is as far we can go via the use of standard limit formula $(1)$. To go further we need to split $x^{2} – \sin^{2}x$ as $(x – \sin x)(x + \sin x)$ and the denominator is split as $x^{3}\cdot x$. For the first part $$\frac{x – \sin x}{x^{3}}$$ we can use an expansion based on equation $(4)$ namely $$\sin x = x – \frac{x^{3}}{6} + o(x^{3})$$

Since you already received good answers, let me add another point of view.

Rewriting the expression

$$A=\frac1{\sin^2(x)}-\frac1{x^2}=\frac{x^2-\sin^2(x)}{x^2\sin^2(x)}$$ and knowing that, close to $0$, $\sin(x)\approx x$, then the denominator looks like $x^4$ that is to say that, if the limit exist the numerator should also be developed at least up to order $x^4$; again, since $\sin(x)\approx x$, then $\sin(x)$ should be developed for at least one more term.

So, using $\sin(x)=x-\frac{x^3}{6}+O\left(x^4\right)$, after simplication, the numerator is just $\frac{x^4}{3}+\cdots$ and the denominator is $x^4+\cdots$; so the limit $\frac 13$.

If we needed more than the limit (say for example how it is approached), more terms would be required. Using $\sin(x)=x-\frac{x^3}{6}+\cdots$ and using it for both numerator and denominator, we should get $$A\approx \frac{x^2-\left(x-\frac{x^3}{6}\right)^2}{x^2 \left(x-\frac{x^3}{6}\right)^2}$$ which, after simplication and long division would give $$A\approx \frac{1}{3}+\frac{x^2}{12}+\cdots$$
Using instead $\sin(x)=x-\frac{x^3}{6}+\frac{x^5}{120}+\cdots$ and using it for both numerator and denominator, and doing the same as before, we should get $$A\approx \frac{1}{3}+\frac{x^2}{15}+\cdots$$ which is slightly different.

To illustrate the above, I suggest you plot on the same graph the three functions $y_1=\frac1{\sin^2(x)}-\frac1{x^2}$ , $y_2=\frac{1}{3}+\frac{x^2} {12}$, $y_3=\frac{1}{3}+\frac{x^2}{15}$ for $-\frac 12 \leq x\leq \frac 12$ . This would be better than a long and tedious speech.

Since $\sin(x)$ is an entire odd function and $\sin(x)\approx x$ in a neighbourhood of the origin, the limit
$$ L=\lim_{x\to 0}\frac{\sin(x)-x}{x^3}=\lim_{x\to 0}\frac{\sin(2x)-2x}{8x^3} $$
must exist, and be equal to
$$ \frac{4L-L}{3} = \frac{1}{3}\lim_{x\to 0}\frac{4\sin(2x)-8\sin(x)}{8x^3} =\frac{1}{3}\lim_{x\to 0}\frac{\sin(x)\cos(x)-\sin(x)}{x^3} $$
so:
$$ L = -\frac{1}{3}\lim_{x\to 0}\frac{1-\cos x}{x^2} = -\frac{1}{6}\lim_{x\to 0}\frac{\sin^2\frac{x}{2}}{\left(\frac{x}{2}\right)^2} = \color{red}{-\frac{1}{6}}.$$