1

I have trouble understanding when it's ok or not to substitute a function with another asymptotic one.

I know that, when $\lim _{x\to x_0}\left(\frac{f\left(x\right)}{g\left(x\right)}\right)=1$ , if I'm in a limit in which $x\to x_0$ , then I can "replace" f(x) with g(x) .

But WHEN can I replace it?

An example that is making me confused:
Consider this limit:

$\lim _{x\to \:0}\left(\frac{sin\left(x\right)-x+2x^5}{3x^3}\right)$

The result should be $\ell =-\frac{1}{18}$ (Wolfram agrees)

But I'm having trouble with this.

First error I noticed was trying to apply the notable limit $\lim _{x\to \:0}\left(\frac{sin\left(x\right)}{x}\right)=1$.

I grouped by x, and I applied the substitution, and I ended with a wrong result ($\ell =0$).

$\lim \:_{x\to \:\:0}\left(\frac{sin\left(x\right)-x+2x^5}{3x^3}\right)=\lim \:_{x\to \:\:0}\left(\frac{x\left(\frac{sin\left(x\right)}{x}-1+\frac{2x^5}{x}\right)}{3x^3}\right)=\lim \:\:_{x\to \:\:\:0}\left(\frac{x\left(\left[1\right]-1+\frac{2x^5}{x}\right)}{3x^3}\right)=...=0$.

Ok, this is wrong.
But why THIS other approach I tried is wrong?

$\lim \:_{x\to \:\:0}\left(\frac{sin\left(x\right)-x+2x^5}{3x^3}\right)$ ⇒ For x→0, sin⁡(x)∼x ⇒ $\lim \:_{x\to \:\:0}\left(\frac{\left[x\right]-x+2x^5}{3x^3}\right)=...=0$

WHEN is it not appropriate to substitute a function with an asymptot? Help.

Fibo
  • 13
  • You can always replace $A$ by $B$ if $A=B$. Thus you can always replace the expression $\lim\limits_{x\to x_0}\dfrac{f(x)}{g(x)}$ by $1$. But this does not mean that you can replace $f(x) $ by $g(x) $. See https://math.stackexchange.com/a/1783818/72031 – Paramanand Singh Jan 26 '18 at 06:16
  • Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Jan 28 '18 at 07:56

2 Answers2

0

You cannot sum this type of relation only multiply. If you wanna study your example you cannot suppress the "rest" behind it because he's too important in the considered limit :

$$ \frac{\sin\left(x\right)-x+2x^2}{3x^3}=\frac{\left(x-\frac{x^3}{6}-x+2x^5+o\left(x^3\right)\right)}{3x^3} \underset{x \rightarrow 0}{\rightarrow}\frac{1}{6 \times 3}=-\frac{1}{18} $$

The thing is here you used $$\sin\left(x\right) \underset{\left(0^{+}\right)}{\sim}x$$ and you summed equivalent. You can substitute when it is a product, not a sum. For example you would have had instead $$ \frac{2\sin\left(x\right)x^2}{3x^3} $$ You could have wrote

$$ \frac{2\sin\left(x\right)x^2}{3x^3} \underset{\left(0^{+}\right)}{\sim}\frac{2xx^2}{3x^3}\underset{x \rightarrow 0}{\rightarrow}\frac{2}{3} $$

Atmos
  • 7,369
  • I've encountered this equation studying from a youtube channel.
    The guy in the video explained why substituting with the notable limit was wrong, and then he proceeded to show 2 ways to solve this equation (with Taylor and with De L'Hopital).

    The problem is the guy didn't explain why you couldn't apply the asymptotic substitution here.

    So, you're telling me that "you can substitute when it is a product, not a sum".
    Is this the general rule?

    So, I can apply substitutions ONLY if the limit consists in a product/division between functions?

     – Fibo Jan 25 '18 at 18:10
  • You can learn it as a real rule which is pretty always true. You cannot SUM it, but with a PRODUCT you can. There is one exception but you will rarely encounter it ( it is linked with integrals and sums but it is pretty rare to have to use it. ) You used $\displaystyle \frac{\sin(x)}{x}$ tends to $1$ near $0$. You can find it easily, it's a product and $\sin(x) \sim x$ hence it is equivalent to $x/x=1$. that's a product, you can do it – Atmos Jan 25 '18 at 18:14
0

This kind of asymptotic substituition is very complicated and Dangerous to be handled for limits, it is better to proceed by Taylor's expansion with little-o or big-O notation.

In this case note that

$$\sin x= x -\frac{x^3}{6}+o(x^3)$$

then

$$\frac{sin\left(x\right)-x+2x^5}{3x^3}=\frac{x -\frac{x^3}{6}+o(x^3)-x+2x^5}{3x^3}=\frac{-\frac{x^3}{6}+o(x^3)}{3x^3}=\frac{-\frac{1}{6}+o(1)}{3}\to-\frac1{18}$$

Thus in this case you can't replace $\sin x \sim x$ because you need higher order terms to obtain the correct limit. In Others case the first order expansion might be sufficient and in others case you will need to expand further. There are not general rules for it.

user
  • 154,566
  • I've encountered this equation studying from a youtube channel. The guy in the video explained why substituting with the notable limit was wrong, and then he proceeded to show 2 ways to solve this equation (with Taylor and with De L'Hopital). The problem is the guy didn't explain why you couldn't apply the asymptotic substitution here. I know I could solve this particular equation with Taylor. More than that, I'm interested in understanding when I can and when I can't apply substitutions. – Fibo Jan 25 '18 at 18:12
  • To decide wheter or not you can apply the substitution you need to know Taylor's expansion, only by this tool you can decide what kind of asymptotic substitution apply but it has to be done very carefully. – user Jan 25 '18 at 18:15