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I was trying to solve a limit with Taylor series today, which I never had a problem with, but now I suddenly don't understand why it's ok to do it in that way.

For example, the limit

$$\begin{align} &\lim_{x \to 0} \frac{x - x^3/6 - \sin{x}}{x^5} \\ &= \lim_{x \to 0} \frac{(x-x^3/6) - (x - x^3/6 + x^5/5! - \text{higher order terms} )}{x^5} \\ &= \lim_{x \to 0} \frac{-x^5/5! + \text{higher order terms}}{x^5} \\ &= \lim_{x \to 0} {\frac{-1}{5!} + \text{higher order terms}} = - \frac{1}{120} \end{align}$$

But why can we say that the sum of higher order terms goes to zero? I understand that each of the terms goes to zero, but why does their infinite sum go to zero? How can I justify that? I think that although the summands are small, their infinite sum can hypothetically still be big?

Lee David Chung Lin
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Jesus
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    This is a good observation. But you don't need an infinity of terms, it suffices to know that the remainder of the series decreases faster than the last term of the development, so becomes negligible. Review the theory of Taylor's series remainders. (Another explanation is that a Taylor series is an entire series, so it has a radius of convergence. And as long as $x$ belongs to that radius, the series converges, and the sum of the high order terms converges to a power of $x$ times a finite constant.) –  Dec 11 '21 at 14:53
  • Thank you, I had a feeling it has something to do with remainders getting smaller. I'll go review it now. – Jesus Dec 11 '21 at 16:06
  • An infinite sum is the limit of partial finite sums... which are all zero here. – Miguel Dec 11 '21 at 16:36
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    Your question is addressed by this answer and this answer. – user21820 Dec 13 '21 at 16:22

2 Answers2

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I would write it $$\begin{align} &\lim_{x \to 0} \frac{x - x^3/6 - \sin{x}}{x^5} \\ &= \lim_{x \to 0} \frac{(x-x^3/6) - (x - x^3/6 + x^5/5! +O(x^6) )}{x^5} \\ &= \lim_{x \to 0} \frac{-x^5/5! + O(x^6)}{x^5} \\ &= \lim_{x \to 0} \left({\frac{-1}{5!} + O(x)}\right) = - \frac{1}{120} \end{align}$$ and say that the first equality is due to "Taylor's theorem".

GEdgar
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We can do this by Taylor's Mean Value Theorem .

For a $k$ times differentiable function at a point $a$.

$f(x)=f(a)+(x-a)\frac{f^(1)(a)}{1!}+(x-a)^{2}\frac{f^{(2)}(a)}{2!}+...+(x-a)^{k}M_{k}$

Where $M_{k}\to 0$ as $x\to a$.

Mr.Gandalf Sauron
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