Taylor & Maclaurin polynomials number of times you should derivate

Question

I'm trying to understand how many times you should derivate the function which you are working with when using Maclaurin or Taylor. To my understanding the more times you do it the more accurate it becomes, but how do you know when it is enough?

For example:

Determine the limit of: 

lim  ( (e^x - (1-2x)^1/2) / x^2 )
x->0

When i look at the answer it says to start like this: https://i.stack.imgur.com/8gmfs.png

How do you know to stop develop the expression when you get to the second derivative, like in the example? Why not keep going to third, forth, fifth, etc?

Answer 1

You can proceed by trial and error. Develop up to a certain degree and plug in the limit. If all terms cancel out, you need to increase the degree.

With $$\frac{e^x-\sqrt{1+2x}}{x^2},$$ we can try up to degree $1$:

$$\frac{1+x+o(x^2)-(1+x)+o(x^2)}{x^2}$$

As all terms vanish, you know that you need more.

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A little trick:

Pretty often the exercises are arranged so that the limit is a finite number, and you can sometimes guess the degree from the numerator or denominator. In the given example, you can guess degree $2$.

Answer 2

Lets looks at a more general Taylor expression of the form

$$f(x)= \frac{a+bx+cx^2}{x^2}$$

In your example, the $a,b$ cancel out, so you end up with $$f(x)=\frac{0+0x+cx^2}{x^2}=c$$

The limit $\lim_{x\to 0}f(x)$ now computes to $c$, as there is no $x$ term left.

HOWEVER: If you continue to develop the expression, you will get a general result like $$f(x)= \frac{a+bx+cx^2+dx^3}{x^2}$$

Again, $a$ and $b$ cancel out, so you end up with $$f(x)=\frac{0+0x+cx^2+dx^3}{x^2}=c+dx$$

Taking the limit now again leads to $c$, as the higher order terms now have a $x$ inside that gets taken care of with the limit.

So, it doesnt matter when you stop, you will always get the same result (as long as your stopping point is high enough).

Look at the demoninator $x^2$ to get a hint, when you have enough terms to be able to stop.