# calculating limits using infinitesimals

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Normally limits are used instead of infinitesimals, but is it possible to calculate limits using infinitesimals?

For example:

$\lim_{x \to 0} \frac{sin(x) - x}{x^3}$

this is usually solved by applying L'Hopital's rule 3 times and the answer is -1/6:

Edited by 113

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This very limit got worked out on:

math.stackexchange.com/questions/2487889/solve-this-limit-using-equivalent-infinitesimals .

However, the method has the tiny disadvantage that in order to compute the coefficients of the Taylor Series of the expression in the numerator of the fraction, which you need to know, you have to first produce the value given by the limit anyway , I suppose by ordinary means without using infinitesimals?! EDIT: well, after all, the coefficients are values of the iterated derivatives of $$\sin$$ at $$0,$$ which can be done symbolically using infinitesimals, so that part is fine.

Btw, it is either L'Hôpital or L'Hospital to you .

Edited by taeto

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Rather than L'Hopital's, one can take the first few terms of $$sin(x)=x-x^3/3! +x^5/5!+...$$ to get -1/3!, as $$x\to 0$$.

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12 hours ago, mathematic said:

Rather than L'Hopital's, one can take the first few terms of sin(x)=xx3/3!+x5/5!+... to get -1/3!, as x0 .

It depends on the view you take of the Taylor series expansion of $$\sin (x)$$ around $$x =0.$$

Maybe you would think of $$\sin(x)$$ as the imaginary part of $$\exp(ix)$$ with $$\exp$$ defined by its well-known series expansion around $$0$$, in which case your explanation is fine. But that might appear quite mysterious to someone who only knows $$\exp$$ as the inverse of the $$\log$$ function. Which is how $$\exp$$ got reasonably explained back when I went to school.

If not, then to get the third coefficient in the Taylor expansion of $$\sin,$$ isn't that just basically just the value of the limit expression in the OP? And then your answer just seems the same as saying that to calculate the limit, you can use the traditional method. Which is not what the OP wanted.

Edited by taeto

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On 1/25/2020 at 2:37 PM, taeto said:

This very limit got worked out on:

math.stackexchange.com/questions/2487889/solve-this-limit-using-equivalent-infinitesimals .

That's interesting, thanks for sharing. The main question is not answered though: " How many terms should I grab to go safe for every case? Why doesn't it suffice to take just the 1st non-zero term? "

Then they work with the limit:

$\lim_{x \to 0} \frac{tan(x) - sin(x)}{x^3}$

The answer is 1/2,  using the limit calculator:

The limit calculator uses L'Hôpital three times and then plugs in the value 0.

What I was asing in my first post, is it possible to plug in an infinitesimal value? Is it possible to calculate limits using infinitesimals?

Edited by 113

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2 hours ago, 113 said:

How many terms should I grab to go safe for every case? Why doesn't it suffice to take just the 1st non-zero term? "

What I was asing in my first post, is it possible to plug in an infinitesimal value? Is it possible to calculate limits using infinitesimals?

You are thinking of a function given as a series expansion with terms ordered from lower to higher orders. But a function in general may not easily be represented in this way, and it may not even be possible. In some cases there maybe does not exist any first non-zero term to just grab.

Formally to find $$\lim_{x\to 0} f(x)$$ using infinitesimals, you should consider the exact value $$f(\Delta x),$$ as you suggest, by plugging in the infinitesimal value $$\Delta x.$$ Once that is done, and if $$f(\Delta x)$$ is not infinite, you have to compute the standard part function st$$(f(\Delta x) )$$ to get the real value of the limit. How to do that in practice is another story.

If you do happen to know a series expansion of your function, then you do not really need any computation, since the answer depends on whether there are non-zero terms with negative powers, in which case the limit does not exist, and otherwise you read off the constant term.

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