# Not getting the same solution when using the rule sin(x)\x=1 on a limit

There is a rule in limits that when $x$ approaches zero: $$\frac{\sin\left(x\right)}{x}=1$$ So I used this rule on the following exercise:

Evaluate $$\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin\left(2x\right)-\tan\left(2x\right)}$$

I substituted $\sin(2x)$ with $2x$ by the following way:

$$\sin\left(2x\right)=\frac{\sin\left(2x\right)}{2x}\cdot 2x=1\cdot 2x=2x \Rightarrow \lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{2x-\tan\left(2x\right)}$$

But according to symbolab: $$\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{2x-\tan\left(2x\right)}=-\frac{1}{16}$$

while $$\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin(2x)-\tan\left(2x\right)}=-\frac{1}{24}$$

## Why am I getting this contradiction?

More over if I susbsitute the following I do get the right answer $$\tan\left(2x\right)=\frac{\sin\left(2x\right)}{\cos\left(2x\right)}=\frac{2x}{\cos\left(2x\right)}\Rightarrow \lim \:_{x\to \:0}\:\frac{x-\sin\left(x\right)}{2x-\frac{2x}{\cos\left(2x\right)}}=-\frac{1}{24}$$

If you want to test yourself: Symbolab with the excersice preloaded

• Because $\sin 2x$ and $2x$ are NOT equal, but you are repeatedly interchanging them. They are approximately equal when $x$ is small, but it makes a huge difference when your numerator is $x-\sin x$ which is much much smaller than $x$. Jan 19, 2016 at 21:33
• @ErickWong I'm not so sure if that is right. It most certainly bothers me, I will say that. As for the OP, I might say that $\sin(x)=x+O(x^3)$ for small $x$, so the error lies in the $x^3$ term, and if there is still error, it would lie in the $x^5$ term etc. Jan 19, 2016 at 21:36
• In the second try, you substituted sin(2x) by 2x twice, in the first try only once. This seems to be the decisive difference. Jan 19, 2016 at 21:41
• To follow up on @ErickWong's comment: the limit you are using amounts to saying that $\sin x = x + o(x)$ around $0$, where $o(x)$ (Landay notation) says "something that is much smaller than $x$." Now, the issue is that $\tan(x) = x + o(x)$ as well... so what actually matters in the end, since you are taking the difference, is exactly what these two $o(x)$ are. To help with an example: Surely, $\lim_{x\to 0} \frac{1+x}{1}=\lim_{x\to 0} \frac{1+2x}{1}=1.$ But to compute the limit of $\frac{(1+x)-(1+2x)}{x}$, you cannot just replace both parentheses in the numerator by $1$... Jan 19, 2016 at 21:52
• @SimpleArt No, the limit certainly does exist and is straightforward to compute. Computational evidence is weak, probably due to your choice of graphing program having limited precision (there is a lot of cancellation in both factors) Jan 19, 2016 at 21:57

The first order terms cancel in nominator and denominator, so the contributions for the limit are in the third order terms. Replace $\sin(x)$ with $x-\frac16x^3+x^5r(x)$ to get the correct result.

Transforming the term gives $$\frac{\cos 2x}{\cos 2x-1}·\frac{x-\sin x}{2\sin x\cos x} = \frac{\cos 2x}{\cos x}·\frac{x-\sin x}{-2\sin^2 x·2\sin x}$$ The limit of the cosines is $1$, in the denominator one can replace $\sin x$ with $x$ and in the numerator the third order formula to get $$\frac{x-(x-\frac16 x^3+O(x^5))}{-4x^3+O(x^5)} = -\frac{1-O(x^2)}{24}$$

$$\lim_{x\to0}\frac{\sin(x)}x=1$$ doesn't allow you to write

$$\frac{\sin(x)}x=1\text{ nor } \sin(x)-x=0\ !$$

There are circumstances where $\sin(x)-x$ cannot be neglected because it is amplified. For the sake of illustration, here is a plot of

$$\frac{\sin(2x)-2x}{\sin(x)-x}.$$

The problem you encountered was that both $\sin(x)-\tan(x)=O\!\left(x^3\right)$ and $x-\sin(x)=O\!\left(x^3\right)$, this means that substituting $x$ for $\sin(x)$ may substantively change the limit.

To be precise, \begin{align} \sin(x)-\tan(x) &=\left(x-\frac{x^3}6+O\!\left(x^5\right)\right)-\left(x+\frac{x^3}3+O\!\left(x^5\right)\right)\\ &=-\frac{x^3}2+O\!\left(x^5\right) \end{align} whereas \begin{align} x-\tan(x) &=x-\left(x+\frac{x^3}3+O\!\left(x^5\right)\right)\\ &=-\frac{x^3}3+O\!\left(x^5\right) \end{align} This explains why the limits you got differ by a factor of $\frac32$.

I think the limit notation has been made so confusing thanks to various educators (including book authors) who try to simplify it too much. When you say $$\lim_{x \to 0}\frac{\sin x}{x} = 1 \tag{1}$$ you don't mean $(\sin x)/x = 1$ when $x \to 0$. Hence it is just not possible to replace $\sin x$ by $x$. However whenever you see the expression $\lim_{x \to 0}\dfrac{\sin x}{x}$ you can replace it by $1$ (because of the equation $(1)$ above).

You can always replace a thing called $A$ by a thing called $B$ when you know that $A$ and $B$ are equal but if $A \neq B$ then you can't replace $A$ by $B$. Unfortunately people are so used to invalid replacements while calculating limits that they don't notice that it is wrong.

The way to solve your limit problem is as follows: \begin{align} L &= \lim_{x \to 0}\frac{x - \sin x}{\sin 2x - \tan 2x}\notag\\ &= \lim_{x \to 0}\dfrac{x - \sin x}{\sin 2x - \dfrac{\sin 2x}{\cos 2x}}\notag\\ &= \lim_{x \to 0}\dfrac{x - \sin x}{\sin 2x}\cdot\frac{\cos 2x}{\cos 2x - 1}\notag\\ &= \lim_{x \to 0}\dfrac{x - \sin x}{\dfrac{\sin 2x}{2x}\cdot 2x}\cdot\frac{\cos 2x}{\cos 2x - 1}\notag\\ &= \lim_{x \to 0}\dfrac{x - \sin x}{2x(\cos 2x - 1)}\cdot\dfrac{1}{\lim_\limits{x \to 0}\dfrac{\sin 2x}{2x}}\cdot \lim_{x \to 0}\cos 2x\notag\\ &= \frac{1}{2}\lim_{x \to 0}\frac{x - \sin x}{x(\cos 2x - 1)}\cdot 1\cdot 1\notag\\ &= \frac{1}{2}\lim_{x \to 0}\frac{\sin x - x}{2x\sin^{2}x}\notag\\ &= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\ &= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\lim_{x \to 0}\frac{x}{\sin x}\cdot\lim_{x \to 0}\frac{x}{\sin x}\notag\\ &= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot 1\cdot 1\notag\\ &= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\notag\\ &= \frac{1}{4}\lim_{x \to 0}\frac{\cos x - 1}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\ &= \frac{1}{12}\lim_{x \to 0}\frac{\cos^{2}x - 1}{x^{2}(1 + \cos x)}\notag\\ &= -\frac{1}{12}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\lim_{x \to 0}\frac{1}{1 + \cos x}\notag\\ &= -\frac{1}{12}\cdot 1\cdot\frac{1}{2}\notag\\ &= -\frac{1}{24}\notag \end{align} The above solution has been presented in a slightly more detailed fashion (than usually necessary for an exam) in order to show how the expression $\lim_{x \to 0}\dfrac{\sin x}{x}$ is replaced by $1$ (and how $\lim_{x \to 0}\cos x$ is replaced by $1$).

I'd like to address the last part of the question, which is understanding why the last pair of substitutions did not alter the limit, even though it was wrong to substitute $\sin x$ for $x$.

On the denominator, you replaced $\sin 2x - \tan 2x$ with $2x - \frac{2x}{\cos 2x}$. Your replacement denominator factors exactly as $$(\sin 2x - \tan 2x) \frac{2x}{\sin 2x},$$

which is why it has the same limiting behavior as the original denominator: you multiplied the denominator by a quantity that converges to exactly $1$, which means you divided the entire expression by a quantity converging to $1$. The limit laws assure that such an operation does not affect the limit.

However, if you had made almost any other substitutions you would have gotten a significantly different result. In other words, it was just by chance that your random manipulation happened to not change the answer :).

When taking limits of an expression you cannot arbritraily replace parts of the expression in isolation. You need to calculate the limit of the entire expression.

In your case you could use l'Hopital three times to get: \begin{align} \lim_{x\to0}\:\frac{x-\sin(x)}{\sin(2x)-\tan(2x)}&=\lim_{x\to0}\:\frac{1-\cos(x)}{2\cos(2x)-2\sec^2(2x)}\\ &=\lim_{x\to0}\:\frac{\sin(x)}{-4\sin(2x)-8\tan(2x)\sec^2(2x)}\\ &=\lim_{x\to0}\:\frac{\cos(x)}{-8\cos(2x)-16\sec^4(2x)-32\tan^2(2x)\sec^2(2x)} \end{align} Now you can simply replace $x$ with $0$ to obtain: $$\lim_{x\to0}\:\frac{\cos(x)}{-8\cos(2x)-16\sec^4(2x)-32\tan^2(2x)\sec^2(2x)}=\frac{1}{-8*1-16*1-32*0*1}=-\frac{1}{24}$$