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Question: $$\lim_{x\to 0}\frac{\tan x-\sin x}{x^3}$$

The answer, by L'Hopital's rule as well as wolfram and desmos is $\frac{1}{2}$

Here's what I did: $$\lim_{x\to0}({\tan x \over x}\times{1\over x^2}-{\sin x \over x}\times{1\over x^2})$$
$$\lim_{x\to0}({1 \over x^2}-{1 \over x^2})=0$$

Im not sure where the mistake is.

Anvit
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    You can break up limits over product, provided both exist. Now when you split it over product, you must put in the values of respective limits. Clearly you have not done that, whence you will get $\infty - \infty$ – jonsno Jan 04 '18 at 14:34
  • @samjoe that answers my question. can you add it as answer? – Anvit Jan 04 '18 at 14:34
  • Yeah my mistake it was supposed to be tanx - sinx not other way round – Anvit Jan 04 '18 at 14:36
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    @Anvit Also refer to StackTD's answer. Cheers! – jonsno Jan 04 '18 at 14:36
  • See: https://www.desmos.com/calculator/o1mfcduwqh and https://www.wolframalpha.com/input/?i=lim+(x+-%3E+0)(tan+x+-sin+x)%2Fx%5E3 – For the love of maths Jan 04 '18 at 14:40
  • @MohammadZuhairKhan Yeah i had already seen those (I even wrote that after the question) – Anvit Jan 04 '18 at 14:41

3 Answers3

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Im not sure where the mistake is.

You're not being careful with how (properties of) limits work.

$$\lim_{x\to0}({\tan x \over x}\times{1\over x^2}-{\sin x \over x}\times{1\over x^2}) \\ \lim_{x\to0}({1 \over x^2}-{1 \over x^2})=0$$

You're skipping a few (dangerous) steps here, starting with:

$$\lim_{x\to0}\left({\tan x \over x}{1\over x^2}-{\sin x \over x}{1\over x^2}\right) \color{red}{=} \lim_{x\to0}\left({\tan x \over x}{1\over x^2}\right)-\lim_{x\to0}\left({\sin x \over x}{1\over x^2}\right)$$ This is only allowed if the two limits in the right-hand side exist, and they don't.

Suppose you do arrive at those two limits, then you still can't do: $$\lim_{x\to0}\left({\tan x \over x}{1\over x^2}\right)-\lim_{x\to0}\left({\sin x \over x}{1\over x^2}\right) \color{red}{=} \left(\lim_{x\to0}\color{blue}{{\tan x \over x}}\right){1\over x^2}-\left(\lim_{x\to0}\color{blue}{{\sin x \over x}}\right){1\over x^2} $$ and only take the limit of the blue functions, leaving the fractions ${1\over x^2}$ to cancel them at the end.

StackTD
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    Oh i get it now. I straight up applied Lim(f(x)g(x))=Lim(f(x))Lim(g(x)) but they should exist. $1/x^2$ doesnt exist so cant apply this rule. – Anvit Jan 04 '18 at 14:39
  • @AnvitGarg you had 2 mistakes, like StackTD said: one is like you said, and the other is you said $\lim (f(x)-g(x))=\lim f(x)-\lim g(x)$ without knowing that $\lim f(x)$ and $\lim g(x)$ exists – ℋolo Jan 04 '18 at 14:41
  • @Holo In this case also both should exist only then it can be applied. Am I right? – Anvit Jan 04 '18 at 14:42
  • @AnvitGarg yes exactly, and in your case both $\lim \sin x/x^3$ and $\lim \tan x/x^3$ doesn't exists, so you can't – ℋolo Jan 04 '18 at 14:44
  • @AnvitGarg Your second mistake is a bit more subtle; you didn't do $\lim\left(fg\right)=\lim f\lim g$, but you did: $\lim\left(fg\right)=\left(\lim f\right) g$ and not take the limit of $g$; doing this twice to then cancel the $g$'s before taking the limit again. Or, because you left out some steps, perhaps you did intend to distribute the limit of the product to leave $\lim g$ twice and then combine them again to $\lim\left(g-g\right)$, which you can't do either. – StackTD Jan 04 '18 at 14:56
  • @StackTD no, $1/x^2$ is still under limit with both sinx and tanx – Anvit Jan 04 '18 at 14:59
  • @AnvitGarg Yes, but in the next step the trig fractions are gone so in those missing steps, either of the two (wrong) scenarios I described above happened. – StackTD Jan 04 '18 at 15:00
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Let’s look at the limit in two ways, using trigonometric properties on one hand and using Taylor expansions on the other. The indetermination is $0/0$ ; what you do is just writing $0/0=(0-0)/0=0/0-0/0$. This is still indeterminate.

Trigonometry

Write

$$\begin{align} {\tan{x}-\sin{x}\over x^3}&={\tan{x}\left(1-\cos{x}\right)\over x^3}\\ &={\tan{x}\over x}\cdot{1-\cos{x}\over x^2} \end{align}$$

The first term of the product $\tan{x}/x\to1$ and we are left with $(1-\cos{x})/x^2$. Now remember that $\cos{x}=1-2\sin^2{t}$ with $t=x/2$ ($t\to 0$ as $x\to 0$). So we have

$${1-\cos{x}\over x^2}=2{\sin^2{t}\over 4t^2}\to {1\over 2}$$

Taylor

One has

$$\begin{align} &\tan{x}=x+{x^3\over3}+o(x^3)\\ &\sin{x}=x-{x^3\over 6}+o(x^3) \end{align}$$

And so

$${\tan{x}-\sin{x}\over x^3}={x^3\left({1\over 3}+{1\over 6}\right)\over x^3}+o(1)={1\over 2}+o(1)$$

marwalix
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Note that there is no split of the limit like $$\lim f(x) - \lim g(x) $$ here. Rather the expression under limit has been split as a difference based on laws of algebra and this is perfect.

The mistake happens in the next step and is very common and that is replacing the expressions $(\sin x) /x$ and $(\tan x) /x$ by $1$. And that's just plain wrong. We all know that these expressions are never equal to $1$ and thus they can't be replaced by $1$. I really find it surprising that the mistake is so common inspite of the very obvious mathematical fact that one can't replace $A$ by $B$ unless $A=B$.

Well what you can really do is that you can always replace the expression $\lim_{x\to 0}\dfrac{\sin x} {x}$ with $1$ without any restrictions simply because they are equal. This emphasizes the fact the expression $\dfrac{\sin x} {x} $ is different from the expression $\lim_{x\to 0}\dfrac{\sin x} {x} $. Unless this simple fact is taken into consideration one can get into trouble.

I have described this problem in detail in this answer which also describes when such replacements are valid.

Paramanand Singh
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