Admittedly, there are some questions and answers as to why a function defined on a closed interval is not differentiable on the interval's endpoints. But I find no answer as to why, in spite of this, continuity can be defined on an interval's endpoints.  For differentiability, the intuition is that the neighborhood of $x$ that allows $\lim_{x \rightarrow c}\frac{f(x) - f(c)}{x - c} = L\quad$ must (or is this our definition) be populated from both the left ($x &lt; c\;$) and the right ($x &gt; c\;$). Hence, differentiability isn't defined on endpoints of an interval.    Why does such a requirement not exist for continuity?  A function $f$ is continuous at $c$ if $$ \forall \epsilon, \exists \delta, |x - c| &lt; \delta \rightarrow |f(x) - f(c)| &lt; \epsilon$$  Continuity seems not to care whether points appear left or right of $c$ -- just that, if they exist in some chosen neighborhood, they satisfy the condition above.    Here is an example of this "dichotomy"
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In calculus of one variable I read:  A function $f$ is differentiable on an open interval if it is differentiable at every number of the interval.  I wonder why in the definition we suppose that the interval is open. This is the case in Rolle theorem, Mean value theorem,etc.     Do we have a notion of a function being differentiable on a closed interval $[a,b]$?
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I am working on a homework problem from Avner Friedman's Advanced Calculus (#1 page 68) which asks     Suppose that $f(x)$ is a continuous function on the interval $[0,\infty)$. Prove that if $\lim_{x\to\infty} f(x)$ exists (as a real number), then $f(x)$ is uniformly continuous on this interval.   Intuitively, this argument makes sense to me. Since the limit of $f(x)$ exists on $[0,\infty)$, we will be able to find a $\delta &gt; |x_0 - x_1|$ and this implies that, for any $\epsilon&gt;0$, we have $\epsilon &gt; |f(x_0) - f(x_1)|$ (independent of the points chosen). I am aware that the condition of uniform continuity requires that $\delta$ can only be a function of $\epsilon$, not $x$.   What information does the existence of a real-valued limit provide that implies $f(x)$ is uniformly continuous on this interval?
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Assume we have a function $f : \mathbb{R} \rightarrow \mathbb{R}$ such that  $f$ is differentiable for all non-zero $x$, $f$ is continuous at $0$ and  $$\lim_{x \uparrow 0} f'(x) = \lim_{x \downarrow 0} f'(x) &lt; \infty. $$  Is it then true that $f$ is also differentiable at $0$?  If we drop the hypothesis that $f$ is continuous, this is not true since the signums function is a simple counterexample. But what if $f$ is continuous? I don't think this is true but I can't find a counterexample...
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Could someone give an example of a ‘very’ discontinuous derivative? I myself can only come up with examples where the derivative is discontinuous at only one point. I am assuming the function is real-valued and defined on a bounded interval.
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If a function $f$ defined on $[a,b]$ and differentiable everywhere, does it mean that its derivative $f'(x)$ is continuous everywhere on $[a,b]$?  My understanding is 'yes'. Because if there were 'a gap' in $f'(x)$, we could integrate back to $f(x)$ and show that where the gap is, there would be a sharp "angle" where $f(x)$ is not differentiable.
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So I understand the definition of uniform continuity, but wanted some suggestions to prove that a function is or isn't uniformly continuous. I have looked ahead and have seen that if a function is continuous on a compact domain, then the function is also uniformly continuous. But for the purposes of the examples below, I cannot use compactness, or the fact that a function is uniformly continuous if it is continuous and has a bounded derivative. Here's one example:  prove $f(x)= \sqrt{x}$ is uniformly continuous on $(0, \infty)$. I have seen this question asked before on this forum, and the respondents used the traditional definition. Ie. they found a delta such that if $|x-y|&lt;\delta$, then $|\sqrt{x}-\sqrt{y}|&lt; \epsilon$. I was wondering if I could try this in a slightly different way. Suppose I prove that $f$ is continuous at $x_0$, and the choice of $\delta$ doesn't depend on $x_0$. After some scratch work, I proved that $f(x)$ is continuous at $x_0$ provided that $|x-x_0|&lt; \min\{1,(\sqrt{x_0+1} + \sqrt{x_0})*\epsilon\}$. Since $x_0$ is always greater than $0$, we can say that $\delta= \min\{1, \epsilon\}$. Because I proved that our choice of $\delta$ doesn't depend on $x_0$, is this a valid proof that $f(x)$ is uniformly continuous? Here's another example: I want to prove that $f(x)=x^2$ is not uniformly continuous on $(0, \infty)$. After some work, I proved that $f(x)$ is continuous at $x_0$ (in dom $f$), provided that $\delta&lt; min \{1, \frac {\epsilon}{ (2x_0+1)}\}$. Here, since our choice of delta does depend on $x_0$, we can say the function is not uniformly continuous. So am I approaching these problems correctly? I realize that I have not been using the definition of uniformly continuity. (that if $f$ is uniformly continuous, then if $|x-y|&lt;\delta$, then$|f(x)-f(y)|&lt; \epsilon$. Also, sorry about the formatting. I realize I should learn proper formatting soon!. Also, I have read the following post regarding this topic:
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How would I give a simple example of a function $f$ differentiable in a deleted neighborhood of $x_0$ such that $\lim_{x\to x_0}f^\prime(x)$ does not exist? I cannot seem to think of an example.  A delete neighborhood is an open interval about $x_0$ which does not contain $x_0$. So, $(x_0-\delta,x_0+\delta)-\{x_0\}$ for some $\delta&gt;0$.  How would something be differentiable in a deleted neighborhood if at the point of the derivative, the limit does not exist. Presumably, the derivative ends up looking something like $\lim_{x\to x_0} \dfrac{1}{x}$, if it does not exist.
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If a function $f$ is continuous at every $c$ point in $[a, b]$ then we have for every $\epsilon$ we can find a $\delta$ such that whenever $|x-c|&lt;\delta$, we have $|f(x)-f(c)|&lt;\epsilon$.      Now, in this definition, $c$ varies in the interval $[a, b]$, and also $x$ varies in $[a, b]$, then what is the difference between $x$ and $c$?
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Using the definition of continuity and the $\epsilon$-$\delta$ definition of a limit, how would you proceed to show that the following statement holds true?  Let $f : \Bbb{R} \rightarrow \Bbb{R}$ be a continuos function at $x=0$, with the property that if $x\neq 0$ then $f(x)\geq 0$. Then $f(0) \geq 0$.  My attempt so far has been to assume that $f(0) \lt 0$, which means that $$f(0)=\lim_{x\to 0} f(x)\lt 0,$$ because of the continuity at the given point. Furthermore, I think this means that when $x$ approaches $0$ from both sides, at some point the value $f(x)\lt 0$, thus forming a contradiction.   I can't seem to see what to do afterwards. Is this a way of approaching the problem? And how will the $\epsilon-\delta$ definition come into play if my approach is usable?
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Suppose we have a continuous function $f : \mathbb{R} \to \mathbb{R}$. Suppose also that for a certain point $c$, $\lim_{x \to c} f'(x)$ exists. Must $f'(c)$ exist as well, and be equal to this limit?  This isn't quite the same as asking if derivatives are always continuous. The well-known function $f(x) = x^2 \sin (1/x)$ is continuous and differentiable everywhere, but its derivative has no limit at $x = 0$. I'm wondering if the derivative of a continuous function can have a discontinuity where its limit does exist.
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As the title indicates, I'd like to prove the following:     If $f:\mathbb R\to\mathbb R$ is a continuous function on $[a,b]$, then $f$ attains its maximum.   Now, I do have a working proof: $[a,b]$ is a connected, compact space, which means that because $f$ is continuous, $f([a,b])$ is compact and connected as well.  Therefore, $f([a,b])$ is a closed interval, which means it has both a minimum and, as desired, a maximum.  What I would like, however, is a proof that doesn't require such general or sophisticated framework.  In particular, I'd like to know if there's a proof that is understandable to somebody beginning calculus, one that (at the very least) doesn't invoke compactness.  Any comments, hints, or solutions are welcome and apreciated.
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Suppose $f(x)$ is continuous on $[a, b]$ except at a point $c$ in $(a, b)$ at which $f(x)$ has a jump discontinuity. For $x$ in $(a,b)$, set $F(x)=\int_a^xf(t) \, dt$. Show that $F(x)$ is continuous but not differentiable at $x =c$.  Hint: first explain why $f(x)$ is integrable on $[a,b]$, then consider one-sided derivatives of $F(x)$ at $x =c$.  I don't know how to step up because it seems conflict to me. $f(x)$ is not continuous because it has a jump but I have to prove it continuous, how? Can someone guide me, please.
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Suppose that $f$ is a function defined in $[a;b]$ to $[a;b]$ and continuous on $[a;b]$. The problem is I haven't the definition of the function, this is more abstract, but even if how can I prove that $f$ would have a fixed point?
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The question I'm trying to answer is:  Let $I$ be an interval and let $f:I\rightarrow\mathbb{R}$ be differentiable on $I$. Show that if the derivative $f'$ is never 0 on $I$, then either $f'(x)&gt;0$ for all $x$ or $f'(x)&lt;0$ for all $x$.  I know I can apply Darboux's Theorem to solve the question.  My Attempt: Let $I=[a,b]$. If $k$ is a number s.t. $f'(a)&lt;k&lt;f'(b)$, then there exists a point $c$ in $(a,b)$ s.t. $f'(c)=k$. Because $f'\neq0$, $0&lt;f'(a)&lt;k&lt;f'(b)$.  I'm really stuck.. Could somebody help me out here? Thank you.
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Assume that $f:[a,b] \to \mathbb R$ is a function.    Prove the following statement :       If $f$ is differentiable on $[a,b]$ and $f'$ is monotonic on $[a,b]$,   then $f'$ is continuous on $[a,b]$.     Note 1 : We've learned a theorem in the class which says :       If $I$ is an open interval and $f:I\to\mathbb R$ is monotonic, then   $f$ is continuous on $I$.   So, here i can say that $f'$ is continuous on $(a,b)$. But my problem is how to prove that $f'$ is also continuous at the end points $a$,$b$.    Note 2 : There's a similar question  but please pay attention that my question is asking about when $f$ is defined on a closed interval.
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This should be a very easy question but it might just be that I'm confusing myself. So we have the definition of a function $f$ on $S$ being continuous at $x_0$:     For any $\epsilon$>0, there exists $\delta&gt;0$ such that: whenever $|x-x_0|&lt;\delta$, we have $|f(x)-f(x_0)|&lt;\epsilon$   And I assume the negation is     There exists $\epsilon$>0 such that for all $\delta&gt;0$, $|x-x_0|&lt;\delta$ yet $|f(x)-f(x_0)|\ge \epsilon$.   Now I want to show that the function $f(x)=\sin(\frac{1}{x})$ together with $f(0)=0$ cannot be made into a continuous function at $x=0$. So I need to show that there exists $\epsilon&gt;0$ such that for all $\delta&gt;0$, $|x|&lt;\delta$ yet $|f(x)|\ge\epsilon$.  Let $\epsilon = \frac{1}{2}$. Then no matter what $\delta$ we choose, let $|x|&lt;\frac{1}{2}$. It is certainly possible that $|f(x)|\ge \frac{1}{2}$, because, well, $\frac{1}{x}$ can really take on arbitrarily large value as $x$ is small.  Now, what confuses me is that, as $x$ gets small, $f(x)$ can certainly be greater than $\frac{1}{2}$ for infinitely many times, but it will be less than that infinitely many times, too. But I suppose it doesn't really matter. So I think there's something wrong with my negation but I couldn't figure out where.  Update: The correct version can be found . Watch for Lemma 4.6
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So here is my proof, which after looking up others seems to be too simple, or not rigourous enough, though I don't see why (hence I am asking!):  We take the contrapositive, and so prove that if $f$ is unbounded on $[a,b]$, then it is not continuous on this interval.  By hypothesis, $\exists c \in [a,b] : \displaystyle\lim_{x\to c} f(x) = \infty \implies f(c)$ is undefined. (or, $ \displaystyle\lim_{x\to c} f(x) \not= f(c)$ so we get discontinuity instantly)  Since $f(c)$ doesn't exist, the definition of continuity cannot be applied so $f$ must be discontinuous at at least $x=c$, as required.