Trigonometric limit mistake

Question

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.

Answer 1

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.

Answer 2

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

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

Answer 3

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.