Error evaluating $ \lim_{x\to 0}\frac{x-\tan x}{x^3} $

Question

Evaluate the limit: $$ \lim_{x\to 0}\frac{x-\tan x}{x^3} $$

I solved it like this,

$$ \lim_{x\to 0} \left({1\over x^2} - \frac{\tan x}{x^3}\right) =\lim_{x\to 0}\left({1\over x^2} - {\tan x\over x}\cdot {1\over x^2}\right) $$

Now using the property $$ \lim_{x\to 0} \frac{\tan x}{x}=1 $$

we have:

$$ \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{x^2}\right)=0 $$

Please explain my error! How can I avoid such errors? I have it's correct solution. All I want to know is what I did wrong here?

Note: English is my second language.

Answer 1

The explanation you're looking for is this. You are implicitly using the properties $\lim\limits (f+g)=\lim\limits(f)+\lim\limits(g)$ and $\lim\limits (fg)=\lim\limits(f)\lim\limits(g)$, but this is only true when all the limits in these equalities exist, (disclaimer: there are important assumptions that I'm not writing but that should accompany these properties). More specifically, what you did (implicitly) was: $$ \begin{align} \lim\limits_{x\to 0} \left({1\over x^2} - \frac{\tan x}{x^3}\right) &=\lim\limits_{x\to 0}\left({1\over x^2} - {\tan x\over x}\cdot {1\over x^2}\right)\\ &=\lim\limits_{x\to 0}\left({1\over x^2}\right) + \lim\limits_{x\to 0}\left(- {\tan x\over x}\cdot {1\over x^2}\right) \tag{Incorrect}\\ &=\lim\limits_{x\to 0}\left({1\over x^2}\right) - \lim\limits_{x\to 0}\left( {\tan x\over x}\right)\lim\limits_{x\to 0}\left({1\over x^2}\right) \tag{Incorrect}\\ &=\lim\limits_{x\to 0}\left({1\over x^2}\right) - \lim\limits_{x\to 0}\left({1\over x^2}\right) \tag{*}\\ &=\lim\limits_{x\to 0}\left({1\over x^2} - {1\over x^2}\right) \tag{Incorrect}\\ &=0 \tag{**} \end{align} $$

$(\text*)\text{ As correct as something meaningless can be}$
$(\text{**})\text{ Actually correct, but it's too late}$

You can't just replace the value of the limit inside without using the above reasoning or something else which will end up not working.

Answer 2

Although, in fact, $\lim_{x\to0}\frac{\tan x}x=1$, you cannot deduce from that that$$\lim_{x\to0}\frac1{x^2}-\frac{\tan x}x\times\frac1{x^2}=\lim_{x\to0}\frac1{x_2}-\frac1{x^2}.$$

In this case, L'Hopital's Rule is the way to go:\begin{align}\lim_{x\to0}\frac{x-\tan x}{x^3}&=\lim_{x\to0}\frac{-\tan^2x}{3x^2}\\&=-\frac13\left(\lim_{x\to0}\frac{\tan x}x\right)^2\\&=-\frac13.\end{align}

Answer 3

Another way of solving the problem is by using series expansion at $x=0$, then you have $\lim_{x \rightarrow 0} \frac{x- \tan{x}}{x^3} = \lim_{x \rightarrow 0} \frac{x-(x + \frac{x^3}{3} + O(x^5))}{x^3} = \lim_{x \rightarrow 0} \frac{-\frac{x^3}{3}+O(x^5)}{x^3} = -\frac{1}{3}$

Answer 4

It has been shown in other answers what exactly went wrong with your approach.

As an alternative consider the Taylor expansion of your function under the limit around $0$. It's known that: $$ \tan x \sim x + {x^3 \over 3} + O(x^5) $$

Note that: $$ \frac{\tan x}{x} \sim {1\over x}\left(x + {x^3 \over 3} + O(x^5)\right) = 1 + {x^2\over 3} + O(x^4) $$

Thus as noted by Yves Daoust: ${\tan x \over x} = 1 + f(x)$.

Using this your limit becomes: $$ \lim_{x\to0} \frac{x - \tan x}{x^3} \sim \lim_{x\to0}\frac{x - x - {x^3\over 3}}{x^3} = -{1\over 3} $$

Answer 5

Everyone here is correctly pointing out that you cannot use $\displaystyle \lim_{x\to 0} \dfrac{\tan x}{x}=1$, but they haven't stated why is that.

Now let me try to explain it to you.

Before proceeding I would like to clear a fact.

Let $f(x)$ and $g(x)$ be two functions such that $\displaystyle \lim_{x\to a}f(x)=l$ and $\displaystyle \lim_{x\to a} g(x)=m$ then,

$$\displaystyle \lim_{x\to a}\left(f(x)\pm g(x)\right) =l\pm m $$

But the converse i.e.

$$l\pm m = \displaystyle \lim_{x\to a}\left(f(x)\pm g(x)\right)\ \ \ ....(1)$$

may not be true.

I have marked this as $(1)$ because I'll be referring to it later.

You did this

$$ \lim_{x\to 0} \left({1\over x^2} - \frac{\tan x}{x^3}\right) =\lim_{x\to 0}\left({1\over x^2} - {\tan x\over x}\cdot {1\over x^2}\right) $$

Everything looks fine uptill here. Now you want to use this property $\displaystyle \lim_{x\to 0} \dfrac{\tan x}{x}=1$

So you will have to split this expression like this.

$$ \lim_{x\to 0}\left({1\over x^2} - {\tan x\over x}\cdot {1\over x^2}\right) $$

$$\implies \lim_{x\to 0}{1\over x^2} - \lim_{x\to 0}\left({\tan x\over x}\cdot {1\over x^2}\right) $$

$$\implies \lim_{x\to 0}{1\over x^2} - \lim_{x\to 0}{\tan x\over x}\cdot\lim_{x\to 0} {1\over x^2} $$

Now you can use $\displaystyle \lim_{x\to 0} \dfrac{\tan x}{x}=1$ to get this

$$ \lim_{x \to 0}\frac{1}{x^2} - \lim_{x \to 0}\frac{1}{x^2}$$

Upon reaching this stage, you did this step

$$ \lim_{x \to 0} \left(\frac{1}{x^2} - \frac{1}{x^2}\right) $$

But this is wrong as I already stated in point (1) above.

Now you'd why is that. It is because

$$\lim_{x\to 0}{\tan x\over x}\cdot\lim_{x\to 0} {1\over x^2} \neq \lim_{x \to 0} \frac{1}{x^2} $$

but

$$\lim_{x\to 0}{\tan x\over x}\cdot\lim_{x\to 0} {1\over x^2} = 1.00000000000..........0001 \cdot \lim_{x \to 0} \frac{1}{x^2} $$

This number is so close to $1$ that we just approximate it to $1$.

Therefore

$$\lim_{x\to 0}{1\over x^2} - \lim_{x\to 0}{\tan x\over x}\cdot\lim_{x\to 0} {1\over x^2} = \lim_{x\to 0} {1\over x^2} - 1.00000000000..........0001 \lim_{x \to 0} \dfrac{1}{x^2} $$

That's why I said in point $(1)$ the converse may not be true.

If you properly evaluate the limit as @Jose Carlos Santos and @roman did. You will find the answer to be $-\dfrac{1}{3}$.

You can clearly note that this answer is negative, because $$1.00000000000..........0001 \lim_{x \to 0} \frac{1}{x^2} > \lim_{x\to 0} {1\over x^2} $$

Answer 6

Right, $$\lim_{x\to0}\frac{\tan x}x=1.$$

But that doesn't mean that you can replace $\dfrac{\tan x}x$ by $1$ inside the limit !

Actually,

$$\frac{\tan x}x=1+f(x)\ne1$$ and the function $f$ can strike back.

---

The "striking back" works like this:

• subtracting $1$ from $\dfrac{\tan x}x$ isolates $f(x)$.

• then dividing by $x^2$ "amplifies" it, giving the term $\dfrac{f(x)}{x^2}$. It turns out that $f(x)$ has a double root at $x=0$, so that the fraction bring a finite contribution, but it could very well have been unbounded.

Answer 7

Note that you can only split limits involving two functions if limits of both functions exists individually (and must be a finite, well defined numbers).

Therefore you first step itself is not allowed while evaluating limits. Moreover you can't replace (tanx)/x by 1 inside limit, you will have to first split the limit, but as I have mentioned it's an incorrect operation and would yield absurd results if ignored.