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I've been going through Spivak's calculus and in it spivak states that for a function $f$ to be continuous on a closed interval $[a,b]$ the function is said to be continuous at the end points if $\lim_\limits{x\to{a^+}}f(x)=f(a)$ And $\lim_\limits{x\to{b^-}}f(x)=b$

However can the same reasoning apply to the derivative of a function i know that limits are based off taking limits from any direction within the domain of the function that is $x$ lies in the domain of $f$ so for example if I had the function $f:[a,b]\longrightarrow \mathbb{R}$ when calculating the derivative at the end point for example at b since f is not defined for points greater than b is it true to say that the derivative of the function at b is the left handed derivative of b That is $$ f'(b)=\lim_{h\to 0^-}\frac{f(b+h)-f(b)}{h} $$ Thanks in advance.

Bernard
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The homeschooler
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2 Answers2

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You are correct, but you also don't need a provisional definitions when considering limits. The rigorous definition of a limit for a function $f:A\subseteq \mathbb{R}\to\mathbb{R}$ at a point $x_0$ only consider the elements of the domain. That is we call $L=\lim_{x\to x_0}f(x)$ if for every $\epsilon > 0$ there is a $\delta >0$ such that if $x\in A$ and $|x-x_0|<\delta$ then $|f(x)-f(x_0)|<\epsilon$. That means you only have to consider how the function approaches for $x$ close $x_0$ in the domain, so in the boundary case, the limit and the right-handed limit are the same as the elements above or below these limit points are not in the domain. Some people may have issues with this definition as it would imply all functions defined on, let's say, the integers are continuous. But further down the line you will see this is consistent with the notion of the subspace topology of a given topological space.

Fiter
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  • Does this mean that an isolated point in a functions domain is continous? – The homeschooler Jan 01 '21 at 00:09
  • @Thehomeschooler Yes, that's right. See this thread for details. Basically, if we state the definition of a limit more carefully, then the function is continuous at an isolated point. – Joe Jan 01 '21 at 00:33
  • @fiter i noticed you missed that out in your definition $x\in A$ and 0$<$$|x-x_0|<\delta$ – The homeschooler Jan 01 '21 at 10:33
  • @Thehomeschooler I believe it's equivalent to only consider x!=x_0 but I think the only thing to consider is that x_0 may not be in the domain. This already implicitly tells you what matters are point arbitrarily near. For example, in the definition of the derivative x_0 is not in the domain because that would give you division by 0. – Fiter Jan 02 '21 at 00:52
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Yes, you can do that. The idea is that this derivative enables you to make sense of what would happen if you were to extend $f$ beyond the interval $[a,b]$. However, you need to be careful when extending this new notion of derivative to the theorems you are familiar with. Just take $f : [0,1] \to \mathbb R$ defined by $f(x) = x$. It has a well-defined derivative on $[0,1]$ equal to $1$ everywhere, but your theorem that says that if $f$ takes its maximum at $x_0$, then $f'(x_0) = 0$, is not true anymore. The theorem becomes "if $f$ takes its maximum at $x_0$, then either $f'(x_0) = 0$ or $x_0$ is at the endpoints of your interval".

Hope that helps,

Patrick Da Silva
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  • sorry but what do you mean by the one sided derivative being used to make sense of the derivative if it was extended beyond $[a,b]$ aren't we assuming the function is bounded in this domain ? – The homeschooler Dec 31 '20 at 23:22
  • @Thehomeschooler: What I meant is that your suggestion of the notion of a derivative on $[a,b]$ is the one that would make sense if there existed a differentiable function $\delta > 0$ and a function $g : ]a-\delta, b+\delta[ \to \mathbb R$ satisfying $g|_{[a,b]} = f$. – Patrick Da Silva Dec 31 '20 at 23:24
  • Spivak introduces right-hand and left-hand derivatives while discussing the derivative of $|x|$ (chapter 9, 3rd ed.). I've noticed some of your questions concern integration, while others cover fundamental concepts of limits. This isn't meant to be rude, but if you aren't already doing so, you might want to begin the book from the first chapter and work your way through. Trying to pick up later concepts when you haven't absorbed the earlier stuff is gonna be rough going. – Ben Dec 31 '20 at 23:45
  • True but in the a level books proofs and details such as limits are not explained so I kind of had to work back to get a deeper understanding – The homeschooler Jan 01 '21 at 00:00
  • Understood and I totally sympathize. I'm finding personally it's easier to build anew from the ground up, than to try to repair the many holes in my own understanding ad hoc. It all depends on your goals, time constraints, etc. – Ben Jan 01 '21 at 00:56