<blockquote>
<ul>
<li><strong>Question:</strong> How do I show that: <span class="math-container">$\lim \limits_{∆θ_i \to 0} \frac{R}{∆θ_i} = 0$</span></li>
</ul>
<p>[...]</p>
<ul>
<li><strong>Question:</strong> How do we show that: <span class="math-container">$\lim \limits_{∆x_i \to 0} \frac{R_2(∆x)}{(∆x)^2} = 0$</span></li>
</ul>
</blockquote>
<p>As you can see from your questions, it must be proved not only that the remainders <span class="math-container">$R_1$</span> or <span class="math-container">$R_2$</span> go to zero, but the <strong>ratio between the remainders and <span class="math-container">$(\Delta x) ^2$</span> or <span class="math-container">$\Delta \theta$</span> goes to zero</strong>.</p>
<p>That is, <strong>the remainder <span class="math-container">$R_n$</span> in Taylor's formula is an infinitesimal of greater</strong> <strong>order with respect to <span class="math-container">$(\Delta x)^n$</span></strong>.  This is the important theorem that must be proved with regard to Taylor's formula.</p>
<p>Intuitively, this means that the remainder <span class="math-container">$R_n$</span>, of a Taylor's polynomial of order <span class="math-container">$n$</span>,<span class="math-container">$^1$</span>  goes to zero more 'quickly' tha <span class="math-container">$(\Delta x)^n$</span>.</p>
<p>Therefore, it is not sufficient to prove that <span class="math-container">$R_n$</span> goes to zero, because it can go to zero 'slowly', so that the ratio <span class="math-container">$\frac {R_n}{\Delta x} \nrightarrow 0$</span>.</p>
<p>This is important, as a Taylor's polynomial is an approximation of the function at a point: for example, intuitively, if when <span class="math-container">$\Delta x$</span> is very small the remainder is very high, we haven't a good approximation.</p>
<p>As you rightly noticed, this ratio, as <span class="math-container">$\Delta x \rightarrow 0$</span>, is the indeterminate form  <span class="math-container">$0/0$</span>. Therefore, the proof <strong>uses L'Hôpital's rule</strong>.</p>
<p>I give you below a proof that assumes that the derivatives of the function are continuos, but a similar proof exists for the general case.<span class="math-container">$^2$</span></p>
<p><span class="math-container">$$***$$</span></p>
<p><strong>THEOREM: TAYLOR'S FORMULA</strong>. -  <em>Let <span class="math-container">$f: \mathbb{R} \rightarrow \mathbb{R}$</span> be a function and the derivatives <span class="math-container">$f^{(n)}$</span> exist and are continuous at a point <span class="math-container">$x_0$</span>.</em></p>
<p><em>We have</em></p>
<p><span class="math-container">$$f(x)=  \sum_{k=0}^{n}\frac { f^{(k)}(x_0)}  {k!}(x-x_o)^k+R_n(x)    \qquad (1)$$</span></p>
<p><em>and</em></p>
<p><span class="math-container">$$\lim _{x\rightarrow x_0} \frac {R_n(x)}{(x- x_0)^n}=0  \qquad (2).$$</span></p>
<p>PROOF.</p>
<p>According to the definition of <span class="math-container">$R_n$</span> we must prove that:</p>
<p><span class="math-container">$$\lim _{x\rightarrow x_0} \frac {f(x) -[f(x_0) + f'(x_0) (x-x_0)+ ...+f^{(n)} (x_0)(x-x_0)^n/n!]}{(x- x_0)^n}=0 \qquad (3)$$</span></p>
<p>We now use l'Hôpital rule. The derivative is respect to <span class="math-container">$x$</span>, so the derivative of <span class="math-container">$f(x_0)$</span> is zero, whereas the derivative of <span class="math-container">$\frac {f^{(n)}(x-x_0) ^n}{n!}$</span> is</p>
<p><span class="math-container">$$ \frac {f^{(n)}(x_0)n (x-x_0) ^{n-1}}{n!}= \frac {f^{(n)}(x_0) (x-x_0) ^{n-1}}{(n-1)!}\qquad (4)$$</span></p>
<p>Therefore, the limit <span class="math-container">$(3)$</span>, having applied l'Hôpital, is the same as</p>
<p><span class="math-container">$$\lim _{x\rightarrow x_0} \frac {f'(x) -[f'(x_0) +  ...+f^{(n)} (x_0)(x-x_0)^{n-1}/(n-1)!]}{n(x- x_0)^{n-1}}. \qquad (5)$$</span></p>
<p>If <span class="math-container">$n&gt;1$</span>, we have again an indeterminate form <span class="math-container">$0/0$</span>. After applying <span class="math-container">$n$</span> times l'Hôpital rule, we have:</p>
<p><span class="math-container">$$\lim _{x\rightarrow x_0} \frac {f^{(n)} (x)-f^{(n)} (x_0)}{n!} \qquad (6).$$</span></p>
<p>This last limit is zero, as <span class="math-container">$f^{(n)} (x)$</span> is continuous, by assumption, at <span class="math-container">$x_0$</span>.</p>
<p><span class="math-container">$\Box$</span></p>
<p>If reading the proof with a generic <span class="math-container">$n$</span> is too cumbersome, you can re-read the method of the proof with <span class="math-container">$n=2$</span>.</p>
<hr />
<p><span class="math-container">$^1$</span> <sup>This result holds for every order <span class="math-container">$n$</span> of the Taylor's polynomial.</sup></p>
<p><span class="math-container">$^2$</span> <sup>You can find this proof and the overall theory of the Taylor's polynomial in a good book of calculus/ undergraduate text of mathematical analysis.</sup></p>

CC BY-SA 4.0
2023-02-27T13:29:26.000Z
This is another great answer @BakerStreet, glad to see that it was something that simply required some extra steps I.e. L'Hpoital, rather than a total misunderstanding of the material! Thanks again for taking the time as usual, and apologies for my delay in responding (Last week was busy).
CormJack
CC BY-SA 4.0
2023-02-27T13:31:20.983Z
I will take what you have given me and see if I can use it to make sense of my method for $\lim \limits_{∆x_i \to 0} \frac{R_2(∆x)}{(∆x)^2} = 0$. As in you have already shown me that it works for the nth case...but is it obvious where my approach expanding $f(x)$ and $P_2(x)$ might have gone wrong?
CormJack
CC BY-SA 4.0
2023-02-27T14:37:12.010Z
You're welcome, and don't mind the delay, it is normal to be busy! And thank you very much for your kindness. You didn't misunderstand the matter,  you just hadn't  seen before the proof of Taylor's formula. What expansions do you refer to? Which equations in your question?
BakerStreet
CC BY-SA 4.0
2023-02-27T14:53:58.497Z
I'll write it out more clearly, would you mind though if I post it as a picture? I know it's "bad form" but it will be an unbelievable hassle to write it out in MathJax hahaha
CormJack
CC BY-SA 4.0
2023-02-27T14:57:49.277Z
Write as picture, no problem. May  I give you an advice? It is useful, when writing, to number the equations, so it is easier to understand which equations one refers to.
BakerStreet
CC BY-SA 4.0
2023-02-27T15:01:34.923Z
Very smart...definitely feels like a Stack Exchange pro tip..I will do that from now!
CormJack
CC BY-SA 4.0
2023-02-27T15:08:10.820Z
Actually, that is the usual habit of mathematicians, otherwise they become crazy to refer to  equations. :-)
BakerStreet
CC BY-SA 4.0
2023-02-27T15:23:50.930Z
You definitely need to be voted as a moderator or something for this sight. Many have the knowledge, but I think you also have the required patience and enthusiasm for it! Do you teach professionally? Are you a TA, or jr professor or anything?
CormJack
CC BY-SA 4.0
2023-02-27T15:33:48.890Z
Yes, I've been contract professor at university for several yesrs, I have been teaching Economics to Law students, who are intelligent students, but  they haven't a great aptitude to Economics :-), so a little bit of patience is required. But they were clever and kind students, really nice.
BakerStreet
CC BY-SA 4.0
2023-02-17T22:28:07.743Z
2023-02-18T00:12:32.220Z
2023-02-18T00:12:32.220Z
BakerStreet
<p>A number of proofs in optimisation use the idea that the remainder term in either the differential or the Taylor Approximation go to zero. For example:</p>
<ol>
<li>Some envelope theorem proofs:.</li>
<li>Necessity and Sufficiency of FOC &amp; SOC in optimisation proofs.</li>
</ol>
<p>I'm struggling to understand the Remainder term going to 0 as explained below.</p>
<p><strong>1) Removing the remainder term from the differential</strong></p>
<p>Used in an <a href="https://youtu.be/FX7kcDvEsJk?t=678" rel="nofollow noreferrer">envelope theorem proof</a>:</p>
<p>Lets say we have a constraint function <span class="math-container">$G(\mathbf{(x(θ), θ)} =b$</span></p>
<p>The differential for <span class="math-container">$G$</span> when <span class="math-container">$θ _ι$</span>  changes is:</p>
<p><span class="math-container">$∆G = \sum_{j=1}^{n}[\frac{\partial{G}}{\partial{x_j}}∆x_j] + \frac{\partial{G}}{\partial{θ_i}}∆θ_i +R = 0$</span>  Where R is our remainder term:</p>
<p>Then dividing through by <span class="math-container">$∆θ_i$</span> and taking the <span class="math-container">$\lim \limits_{∆θ_i \to 0}$</span> we get:</p>
<p><span class="math-container">$\lim \limits_{∆θ_i \to 0} \frac{∆G}{∆θ_i} = \lim \limits_{∆θ_i \to 0} \sum_{j=1}^{n}[\frac{\partial{G}}{\partial{x_j}}\frac{∆x_j}{∆θ_i}] + \frac{\partial{G}}{\partial{θ_i}} +\frac{R}{∆θ_i} = 0$</span></p>
<ul>
<li><strong>Question:</strong> How do I show that: <span class="math-container">$\lim \limits_{∆θ_i \to 0} \frac{R}{∆θ_i} = 0$</span></li>
</ul>
<p>I understand why R would go to 0. Because as the change approaches 0 then <span class="math-container">$f(x) = P(x)$</span> Where <span class="math-container">$P(x)$</span> is our Taylor Polynomial, and hence the difference <span class="math-container">$R(x)$</span> goes to 0.</p>
<p>But then don't we have <span class="math-container">$\frac{0}{0}$</span> and i don't know how to resolve this in this case?</p>
<p><strong>2) Removing the remainder term from the second order Taylor Polynomial</strong></p>
<p>I've seen this written as <a href="https://youtu.be/APEct8maDxw?t=1515" rel="nofollow noreferrer">Taylors Theorem</a> I've seen this be used to prove the <a href="https://youtu.be/iPuMrI8wAl4?t=713" rel="nofollow noreferrer">necessity and sufficiency</a> of FOC and SOC.</p>
<p>If we define <span class="math-container">$R_2(∆x) = f(x) - P_2(x)$</span> where <span class="math-container">$P_2(x)$</span> is the second order Taylor polynomial around <span class="math-container">$a$</span>.</p>
<ul>
<li><strong>Question:</strong> How do we show that: <span class="math-container">$\lim \limits_{∆x_i \to 0} \frac{R_2(∆x)}{(∆x)^2} = 0$</span></li>
</ul>
<p>Defining <span class="math-container">$∆x = x - a$</span> My best effort so far from expanding <span class="math-container">$f(x)$</span> and <span class="math-container">$P_2(x)$</span>, we get:</p>
<p><span class="math-container">$\lim \limits_{∆x_i \to 0} [\frac{f'(a)}{∆x} -  \frac{f'(a)}{∆x} - \frac{1}{2}f''(a)]$</span>  = <span class="math-container">$- \frac{1}{2}f''(a)$</span> = <span class="math-container">$- \frac{1}{2}f''(x) ≠ 0$</span></p>
<p>If anyone can help with these it would solve many riddles for me. Expanding to <span class="math-container">$\mathbb{R^n}$</span> and other generalised cases I should then be able to do myself, thanks!</p>

<mathematical-economics><optimization>
Remainder term in Linear Approximations going to 0
CormJack
# Remainder term in Linear Approximations going to 0

(Asked by: username_0 on 2023-02-16)

A number of proofs in optimisation use the idea that the remainder term in either the differential or the Taylor Approximation go to zero. For example:

1. Some envelope theorem proofs:.
2. Necessity and Sufficiency of FOC & SOC in optimisation proofs.

I'm struggling to understand the Remainder term going to 0 as explained below.

**1) Removing the remainder term from the differential**

Used in an [envelope theorem proof](https://youtu.be/FX7kcDvEsJk?t=678):

Lets say we have a constraint function $G(\mathbf{(x(θ), θ)} =b$

The differential for $G$ when $θ \_ι$  changes is:

$∆G = \sum\_{j=1}^{n}[\frac{\partial{G}}{\partial{x\_j}}∆x\_j] + \frac{\partial{G}}{\partial{θ\_i}}∆θ\_i +R = 0$ Where R is our remainder term:

Then dividing through by $∆θ\_i$ and taking the $\lim \limits\_{∆θ\_i \to 0}$ we get:

$\lim \limits\_{∆θ\_i \to 0} \frac{∆G}{∆θ\_i} = \lim \limits\_{∆θ\_i \to 0} \sum\_{j=1}^{n}[\frac{\partial{G}}{\partial{x\_j}}\frac{∆x\_j}{∆θ\_i}] + \frac{\partial{G}}{\partial{θ\_i}} +\frac{R}{∆θ\_i} = 0$

* **Question:** How do I show that: $\lim \limits\_{∆θ\_i \to 0} \frac{R}{∆θ\_i} = 0$

I understand why R would go to 0. Because as the change approaches 0 then $f(x) = P(x)$ Where $P(x)$ is our Taylor Polynomial, and hence the difference $R(x)$ goes to 0.

But then don't we have $\frac{0}{0}$ and i don't know how to resolve this in this case?

**2) Removing the remainder term from the second order Taylor Polynomial**

I've seen this written as [Taylors Theorem](https://youtu.be/APEct8maDxw?t=1515) I've seen this be used to prove the [necessity and sufficiency](https://youtu.be/iPuMrI8wAl4?t=713) of FOC and SOC.

If we define $R\_2(∆x) = f(x) - P\_2(x)$ where $P\_2(x)$ is the second order Taylor polynomial around $a$.

* **Question:** How do we show that: $\lim \limits\_{∆x\_i \to 0} \frac{R\_2(∆x)}{(∆x)^2} = 0$

Defining $∆x = x - a$ My best effort so far from expanding $f(x)$ and $P\_2(x)$, we get:

$\lim \limits\_{∆x\_i \to 0} [\frac{f'(a)}{∆x} - \frac{f'(a)}{∆x} - \frac{1}{2}f''(a)]$ = $- \frac{1}{2}f''(a)$ = $- \frac{1}{2}f''(x) ≠ 0$

If anyone can help with these it would solve many riddles for me. Expanding to $\mathbb{R^n}$ and other generalised cases I should then be able to do myself, thanks!

**Tags:** mathematical-economics, optimization

---

## 1 answers

### Answer by username_1 on 2023-02-17. Score: 2 | Accepted answer

> * **Question:** How do I show that: $\lim \limits\_{∆θ\_i \to 0} \frac{R}{∆θ\_i} = 0$
>
> [...]
>
> * **Question:** How do we show that: $\lim \limits\_{∆x\_i \to 0} \frac{R\_2(∆x)}{(∆x)^2} = 0$

As you can see from your questions, it must be proved not only that the remainders $R\_1$ or $R\_2$ go to zero, but the **ratio between the remainders and $(\Delta x) ^2$ or $\Delta \theta$ goes to zero**.

That is, **the remainder $R\_n$ in Taylor's formula is an infinitesimal of greater** **order with respect to $(\Delta x)^n$**. This is the important theorem that must be proved with regard to Taylor's formula.

Intuitively, this means that the remainder $R\_n$, of a Taylor's polynomial of order $n$,$^1$ goes to zero more 'quickly' tha $(\Delta x)^n$.

Therefore, it is not sufficient to prove that $R\_n$ goes to zero, because it can go to zero 'slowly', so that the ratio $\frac {R\_n}{\Delta x} \nrightarrow 0$.

This is important, as a Taylor's polynomial is an approximation of the function at a point: for example, intuitively, if when $\Delta x$ is very small the remainder is very high, we haven't a good approximation.

As you rightly noticed, this ratio, as $\Delta x \rightarrow 0$, is the indeterminate form $0/0$. Therefore, the proof **uses L'Hôpital's rule**.

I give you below a proof that assumes that the derivatives of the function are continuos, but a similar proof exists for the general case.$^2$

$$\*\*\*$$

**THEOREM: TAYLOR'S FORMULA**. - *Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a function and the derivatives $f^{(n)}$ exist and are continuous at a point $x\_0$.*

*We have*

$$f(x)= \sum\_{k=0}^{n}\frac { f^{(k)}(x\_0)} {k!}(x-x\_o)^k+R\_n(x) \qquad (1)$$

*and*

$$\lim \_{x\rightarrow x\_0} \frac {R\_n(x)}{(x- x\_0)^n}=0 \qquad (2).$$

PROOF.

According to the definition of $R\_n$ we must prove that:

$$\lim \_{x\rightarrow x\_0} \frac {f(x) -[f(x\_0) + f'(x\_0) (x-x\_0)+ ...+f^{(n)} (x\_0)(x-x\_0)^n/n!]}{(x- x\_0)^n}=0 \qquad (3)$$

We now use l'Hôpital rule. The derivative is respect to $x$, so the derivative of $f(x\_0)$ is zero, whereas the derivative of $\frac {f^{(n)}(x-x\_0) ^n}{n!}$ is

$$ \frac {f^{(n)}(x\_0)n (x-x\_0) ^{n-1}}{n!}= \frac {f^{(n)}(x\_0) (x-x\_0) ^{n-1}}{(n-1)!}\qquad (4)$$

Therefore, the limit $(3)$, having applied l'Hôpital, is the same as

$$\lim \_{x\rightarrow x\_0} \frac {f'(x) -[f'(x\_0) + ...+f^{(n)} (x\_0)(x-x\_0)^{n-1}/(n-1)!]}{n(x- x\_0)^{n-1}}. \qquad (5)$$

If $n>1$, we have again an indeterminate form $0/0$. After applying $n$ times l'Hôpital rule, we have:

$$\lim \_{x\rightarrow x\_0} \frac {f^{(n)} (x)-f^{(n)} (x\_0)}{n!} \qquad (6).$$

This last limit is zero, as $f^{(n)} (x)$ is continuous, by assumption, at $x\_0$.

$\Box$

If reading the proof with a generic $n$ is too cumbersome, you can re-read the method of the proof with $n=2$.

---

$^1$ This result holds for every order $n$ of the Taylor's polynomial.

$^2$ You can find this proof and the overall theory of the Taylor's polynomial in a good book of calculus/ undergraduate text of mathematical analysis.

> **Comments:**
> **username_0**: This is another great answer @BakerStreet, glad to see that it was something that simply required some extra steps I.e. L'Hpoital, rather than a total misunderstanding of the material! Thanks again for taking the time as usual, and apologies for my delay in responding (Last week was busy).  
> **username_0**: I will take what you have given me and see if I can use it to make sense of my method for $\lim \limits_{∆x_i \to 0} \frac{R_2(∆x)}{(∆x)^2} = 0$. As in you have already shown me that it works for the nth case...but is it obvious where my approach expanding $f(x)$ and $P_2(x)$ might have gone wrong?  
> **username_1**: You're welcome, and don't mind the delay, it is normal to be busy! And thank you very much for your kindness. You didn't misunderstand the matter,  you just hadn't  seen before the proof of Taylor's formula. What expansions do you refer to? Which equations in your question?  
> **username_0**: I'll write it out more clearly, would you mind though if I post it as a picture? I know it's "bad form" but it will be an unbelievable hassle to write it out in MathJax hahaha  
> **username_1**: Write as picture, no problem. May  I give you an advice? It is useful, when writing, to number the equations, so it is easier to understand which equations one refers to.  
> **username_0**: Very smart...definitely feels like a Stack Exchange pro tip..I will do that from now!  
> **username_1**: Actually, that is the usual habit of mathematicians, otherwise they become crazy to refer to  equations. :-)  
> **username_0**: You definitely need to be voted as a moderator or something for this sight. Many have the knowledge, but I think you also have the required patience and enthusiasm for it! Do you teach professionally? Are you a TA, or jr professor or anything?  
> **username_1**: Yes, I've been contract professor at university for several yesrs, I have been teaching Economics to Law students, who are intelligent students, but  they haven't a great aptitude to Economics :-), so a little bit of patience is required. But they were clever and kind students, really nice.  

---

