Suppose that $f$ is continuous at $a$, and that $f'(x)$ exists for all $x$ in some interval containing $a$, except perhaps for $x=a$. Suppose, moreover, that $\lim \limits_{x \to a} f'(x)$ exists. Then $f'(a)$ also exists, and $$f'(a) = \lim \limits_{x \to a} f'(x).$$  Proof  By definition, $$f'(a) = \lim \limits_{x \to a} \frac{f(a+h) - f(a)}{h}.$$  For sufficiently small $h &gt; 0$ the function $f$ will be continuous on $[a, a + h]$ and differentiable on $(a, a + h)$ (a similar assertion holds for sufficiently small $h &lt; 0$). By the mean value theorem there is a number $\alpha_h$ in $(a, a + h)$ such that $$\frac{f(a + h) -f(a)}{h} = f'(\alpha_h).$$  Now $\alpha_h$ approaches $a$ as $h$ approaches $0$, because $\alpha_h$ is in $(a, a + h)$; since $\lim \limits_{x \to a} f'(x)$ exists, it follows that $$f'(a) = \lim \limits_{h \to 0} \frac{f(a + h) - f(a)}{h} = \lim \limits_{h \to 0} f'(\alpha_h) = \lim \limits_{x \to a} f'(x).$$  The suggestion is to apply a rigorous argument to the last step. To show that $\lim \limits_{h \to 0} \alpha_h = a$ it suffices to choose $\delta = \epsilon$ as then if $|h| &lt; \delta$, then $|\alpha_h - a| &lt; \delta = \epsilon$.   If $\lim \limits_{x \to a} f'(x) = l$, then for $\epsilon &gt; 0$ there is a $\delta &gt; 0$ such that if $|x-a| &lt; \delta$, then $|f'(x) - l| &lt; \epsilon$. Thus if $|h| &lt; \delta$, then $|\alpha_h - a| \leq | (a + h) - a| &lt; \delta$ so $|f'(\alpha_h) - l| &lt; \epsilon$. Thus $\lim \limits_{h \to 0} f'(\alpha_h) = l$.   Is this correct?
10069
Let $f$ be a real-valued function continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose that $\lim_{x\rightarrow a}f'(x)$ exists. Then, prove that $f$ is differentiable at $a$ and $f'(a)=\lim_{x\rightarrow a}f'(x)$.    It seems like an easy example, but a little bit tricky. I'm not sure which theorems should be used in here.    ==============================================================  Using @David Mitra's advice and @Pete L. Clark's notes I tried to solve this proof. I want to know my proof is correct or not.  By MVT, for $h&gt;0$ and $c_h \in (a,a+h)$ $$\frac{f(a+h)-f(a)}{h}=f'(c_h)$$ and $\lim_{h \rightarrow 0^+}c_h=a$.    Then $$\lim_{h \rightarrow 0^+}\frac{f(a+h)-f(a)}{h}=\lim_{h \rightarrow 0^+}f'(c_h)=\lim_{h \rightarrow 0^+}f'(a)$$   But that's enough? I think I should show something more, but don't know what it is.
99902
Let $f$ be a continously differentiable function on $\mathbb R$.  Suppose that   $$L=\lim_{x\to\infty}(f(x)+f'(x))$$ exists.  If $0&lt;L&lt;\infty$, then which of the following statements is\are correct?               If $\lim_{x\to\infty} f'(x)$ exists, then it is $0$.     If $\lim_{x\to\infty} f(x)$ exists, then it is $L$.     If $\lim_{x\to\infty} f'(x)$ exists, then $\lim_{x\to\infty}f(x)=0$.     If $\lim_{x\to\infty} f(x)$ exists, then $\lim_{x\to\infty}f'(x)=0$.           My Guess  I could not conclude the answer and prove that properly. But, I guess that it must be 1 and 2. help me.
104239
I need to  Let $f : (a, b) → R$ be differentiable on $(a, b)$. Prove that $f$ is continuous on $(a, b)$ by showing that:  $∀x_0 ∈ (a, b)$ $∀ε &gt; 0$ $∃δ &gt; 0$ s.t.  $(x ∈ (a, b), |x − x_0| &lt; δ) ⇒ |f(x) − f(x_0)| &lt; ε$  My understanding so far:   When we say a function is differentiable at $x_0$, we mean that the limit: $$‎f^{\prime} ‎(x) = \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0}$$ exists.  When we say a function is continuous at $x_0$, we mean that: $$\lim_{x\to x_0} f(x) - f(x_0) = 0$$  My attempted proof:  Let us suppose that $f$ is differentiable at $x_0$. Then $$ \lim_{x\to x_0} \frac{f(x) - f(x_0)}{x-x_0} =  ‎f^{\prime} ‎(x) $$  and hence  $$ \lim_{x\to x_0} f(x) - f(x_0) = \lim_{x\to x_0} \left[ \frac{f(x) - f(x_0)}{x-x_0} \right] \cdot \lim_{x\to x_0} (x-x_0) = 0$$  My issue:  My issue may be a non issue, however for the 6 marks this question should be worth my proof seems shallow. Could someone please poke some holes in my proof, or add to it?   Thank you
96450
Just wanted to confirm that this is a correct solution:    Proof: Suppose $f(x_0) &gt; 0$ for some $x_0 \in [a,b]$. Then, by continuity of $f$, for $\epsilon &lt; f(x_0)$, there exists $\delta &gt; 0$ so that, if $|x-x_0| &lt; \delta$, we have $|f(x) - f(x_0)| &lt; \epsilon$. That is to say, $f(x)$ is non-zero in a neighborhood $B_\delta (x_0)$. Now,   \begin{eqnarray*} U(f,P) &amp;=&amp; \sum_{P([a,b])} M_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} M_i \Delta_i \\ &amp;=&amp; \sup_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \sup_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \sup_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;&gt;&amp; 0 \end{eqnarray*}  In the same manner,   \begin{eqnarray*} L(f,P) &amp;=&amp; \sum_{P([a,b])} m_i \Delta_i \\ &amp;=&amp; \sum_{P(B_\delta(x_0))} m_i \Delta_i \\ &amp;=&amp; \inf_{x \in [x_0-\delta, c_1]} |f(x)| (c_1-(x_0 - \delta)) + \inf_{x \in [c_1,c_2]} |f(x)| (c_2-c_1) + \cdots + \inf_{x \in [c_n, x_0 + \delta]} |f(x)| ((x_0+\delta)- c_n) \\ &amp;\geq&amp; 0 \end{eqnarray*}    We can conclude from here that $\int_{a}^{b} f(x) dx &gt; 0$.  Is this a valid proof?
84598
Let $f$ be a function defined in the open interval $(a,b)$ and let $x_0\in(a,b)$. Suppose in addition that $f'(x)$ exists for all $x_0\neq x\in(a,b)$. Is the following statement true:      If $\lim\limits_{x\to x_0}f'(x)$ exists, then $f'(x_0)$ exists and $\lim\limits_{x\to x_0}f'(x)=f'(x_0)$.   Thanks!
84595
If we let $f_n$ be a sequence of functions on $[a,b]$ such that,  -there exists a point $x_0\in[a,b]$ where $lim_{n-&gt;\infty}f_n(x_0)$ exists -each $f_n$ is differentiable -the sequence $(f'_n)$ converges uniformly on $[a,b]$  I want to prove that $f_n$ converges uniformly on $[a,b]$ to some function $f$  My logic:  if I use the triangle inequality to write,  $$|f_n(x)-f_m(x)|&lt;=|(f_n(x)-f_m(x))-(f_n(x_0)-f_m(x_0))|+|f_n(x_0)-f_m(x_0)|$$  Then I can try and apply the mean value theorem to the first term on the left hand side to prove that $f_n$ converges uniformly on $[a,b]$ to some function $f$.  The mean value theorem states,  Suppose f(x) is a function that satisfies both of the following conditions,  -$f(x)$ is continuous on the closed interval $[a,b]$ -$f(x)$ is differentiable on the open interbal $(a,b)$  Then there is a number $c$ such that $a&lt;c&lt;b$ and,  $f'(c)=\frac{f(b)-f(a)}{b-a}$
49415
I've been thinking about this problem: Let $f: (a, +\infty) \to \mathbb{R}$ be a differentiable function such that $\lim\limits_{x \to +\infty} f(x) = L &lt; \infty$. Then must it be the case that $\lim\limits_{x\to +\infty}f'(x) = 0$?  It looks like it's true, but I haven't managed to work out a proof. I came up with this, but it's pretty sketchy:  $$ \begin{align} \lim_{x \to +\infty} f'(x) &amp;= \lim_{x \to +\infty} \lim_{h \to 0} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \lim_{x \to +\infty} \frac{f(x+h)-f(x)}{h} \\ &amp;= \lim_{h \to 0} \frac1{h} \lim_{x \to +\infty}[f(x+h)-f(x)] \\ &amp;= \lim_{h \to 0} \frac1{h}(L-L) \\ &amp;= \lim_{h \to 0} \frac{0}{h} \\ &amp;= 0 \end{align} $$  In particular, I don't think I can swap the order of the limits just like that. Is this correct, and if it isn't, how can we prove the statement? I know there is  already, but I think this is different in two aspects. First, that question assumes that $\lim\limits_{x \to +\infty}f'(x)$ exists, which I don't. Second, I also wanted to know if interchanging limits is a valid operation in this case.
2881
Let $S = [0,1) \cup [2,3]$ and $f\colon S \rightarrow \mathbb R$ be such that $f(S)$ is connected . Which of the following are true:  a) $f$ is discontinuous exactly at one point.  b) $f$ is discontinuous exactly at two points.  c) $f$ is discontinuous at infinite many points.  d) $f$ is continous.  I think $f$ is continuous . We need to show that $\lim_{x \to a^+}f(x) = f(a)$ for any $a \in S$. Since $f(x)$ is increasing, $\lim_{x \to a^+}f(x) \leq f(a)$. Suppose $\lim_{x \to a^+}f(x) &lt; f(a)$. Then there is $l$ such that $\lim_{x \to a^+}f(x) &lt; l &lt;f(a)$, which contradicts connectedness of $f(S)$. Similarly we can show that $\lim_{x \to a^-}f(x) = f(a)$.  I would be thankful if anyone would check my solution and correct me if there are any errors.
2416
How to prove the following theorem?      Let $I \in R$ be an open interval, let $c \in I$, and let $f: I-{c} \rightarrow R$ be a function. Then $\lim \limits_{x \rightarrow c} {f(x)}$ exists if for each $\epsilon &gt; 0$, there is some $\delta &gt; 0$ such that $x,y \in I$ and $\vert x-c \vert &lt; \delta$ and $\vert y-c \vert &lt; \delta$ implies $\vert f(x)-f(y) \vert &lt; \epsilon$.   I think this needs standard limit definition, sup and inf properties to prove. And I came up with a following scratch of proof:  (1) Suppose for each $\epsilon &gt;0$, there is some $\delta &gt;0$ such that $\vert x-c \vert &lt; \delta$ and $\vert y-c \vert &lt; \delta$ implies $\vert f(x)-f(y) \vert &lt; \epsilon$. For each $r&gt;0$, let $A_r$ = $I \bigcap (c-r,c+r)$. Then, for each $\epsilon&gt;0$, there is some $\delta&gt;0$, such that $x,y \in A_\delta$ implies $\vert f(x)-f(y) \vert &lt; \epsilon$. Then I want to show there is some $a&gt;0$ such that $f(A_a)$ is bounded.  (2) If $f(A_a)$ is bounded, then for each $s \in (0,a)$, $f(A_s) \subseteq f(A_a)$, and thus $f(A_s)$ is bounded. Define $a_s = \mathrm{glb}~f(A_s)$ and $b_s = \mathrm{lub}~f(A_s)$. Let $A = \{a_s \mid s \in (0,a)\}$ and $B=\{b_s \mid s \in (0,a)\}$. Then we know that A has a least upper bound and B has a greatest lower bound, and $\mathrm{lub}(A) \leq \mathrm{glb}(B)$. Now I want to show $\mathrm{lub}(A) = \mathrm{glb}(B)$  (3) If I could show $\mathrm{lub}(A) = \mathrm{glb}(B)$, let $M=\mathrm{lub}(A) = \mathrm{glb}(B)$, I want to show $\lim \limits_{x \rightarrow c} {f(x)} = M$.  Can someone give me some help on how to prove (1) (2) (3), $f(A_a)$ is bounded, $\mathrm{lub}(A) = \mathrm{glb}(B)$, and $\lim \limits_{x \rightarrow c} {f(x)} = M$? Thanks!
78312
Is there a counterexample?  Proposition: Let $f:\mathbb{R} \longrightarrow \mathbb{R}$ be a differentiable function  such that $f':\mathbb{R} \longrightarrow \mathbb{R}$ is differentiable at $a\in\mathbb{R}$ (possibly differentiable at a single point). Then $f':\mathbb{R} \longrightarrow \mathbb{R}$ is continuous at an interval $(a-\delta,a+\delta)$ for some $\delta&gt;0$.  Any hints would be appreciated.
100837
Problem $f(x)$ is defined over $[a,b]$ and differentiable over $(a,b)$, where $b-a\geq 4.$Prove that there exists $\xi \in (a,b)$ such that $f'(\xi)&lt;1+f^2(\xi)$. My Proof Since $b-a \geq 4$，we can obtain $$\exists x_1,x_2 \in (a,b):x_2-x_1&gt;\pi.$$ Denote$$F(x):=\arctan f(x).$$ Obviously, $F(x)$ is continuous over $[x_1,x_2]$ and differentiable over $(x_1,x_2)$. Thus, By Lagrange's Mean Value Theorem，$$\exists \xi \in (x_1,x_2) \subset (a,b):F(x_2)-F(x_1)=F'(\xi)(x_2-x_1).$$ Further, $$\frac{f'(\xi)}{1+f^2(\xi)}=F'(\xi)=\frac{F(x_2)-F(x_1)}{x_2-x_1}\leq \frac{|F(x_2)|+|F(x_1)|}{x_2-x_1}&lt;\frac{\frac{\pi}{2}+\frac{\pi}{2}}{\pi}=1,$$ Which implies $$f'(\xi)&lt;1+f^2(\xi).$$ AM I RIGHT? HOPE TO SEE OTHER PROOFS. THX.
15196
In my introductory Analysis course, we learned two definitions of continuity. $(1)$ A function $f:E \to \mathbb{C}$ is continuous at $a$ if every sequence $(z_n) \in E$ such that $z_n \to a$ satisfies $f(z_n) \to f(a)$. $(2)$ A function $f:E \to \mathbb{C}$ is continuous at $a$ if $\forall \varepsilon&gt;0, \exists \delta &gt;0:\forall z \in E, |z-a|&lt;\delta \implies |f(z)-f(a)|&lt;\varepsilon$. The implication $(2)\implies(1)$ is trivial (though I will happily post a proof if there is sufficient interest).  The proof of the implication $(1)\implies(2)$ is worth remarking on, though. Proof that $(1)\implies(2)$: Suppose on the contrary that $\exists \varepsilon&gt;0:\forall \delta&gt;0, \exists z \in E:\left (|z-a|&lt;\delta \; \mathrm{and} \; |f(z)-f(a)|\ge \varepsilon\right )$.  Let $A_n$ be the set $\{z\in E:|z-a|&lt;\frac{1}{n} \; \mathrm{ and }\; |f(z)-f(a)|\ge\varepsilon\}$.  Now use the Axiom of Choice to construct a sequence $(z_n)$ with $z_n \in A_n \; \forall n \in \mathbb{N}$.  But now $a-\frac{1}{n}&lt;z_n&lt;a+\frac{1}{n}\; \forall n \in \mathbb{N}$ so $z_n \to a$.  So $f(z_n) \to f(a)$.  But $|f(z_n)-f(a)|\ge\varepsilon\; \forall n \in \mathbb{N}$, which is a contradiction. You will have noticed that the above proof uses the Axiom of Choice (the lecturer didn't explicitly spell out the dependence, but it's definitely there).  My question is: is it possible to prove that $(1) \implies (2)$ without using the Axiom of Choice.  I strongly suspect that it isn't.  In that case, can anyone prove that we have to use the Axiom of Choice?  I can think of three ways to do this: (A) Show that $\left( (1) \implies (2)\right)\implies \mathrm{AC}$.  I suspect that this statement is untrue. This is definitely untrue, as Arthur points out, because I only used the axiom of countable choice, which is strictly weaker than AC. (B) Show that $(1)\implies (2)$ is equivalent to some other statement known to require the Axiom of Choice (the obvious example being the well-ordering of the real numbers). (C) Construct or show the existence of a model of ZF in which there exist sequences which satisfy $(1)$ but not $(2)$. Of course, if anyone can think of another way, I would be very interested to hear about it. One final note - I am aware that very many theorems in Analysis use the Axiom of Choice in one way or another, and that this is just one example of such a theorem.  If there exists a model of ZF like the one described in (C), is the study of Analysis in that model interesting?
29377
Let $f$ be differentiable and let $\lim_{x \to \infty}f'(x)=+\infty$ prove that: 1) $\lim_{x \to \infty}(f(x)-f(x-1))=+\infty$ and 2) $\lim_{x \to \infty}f(x)=+\infty$.  1) I'll prove by contradiction. Let $\lim_{x \to \infty}(f(x)-f(x-1))=a$, where $a$ is from and $\Bbb R$. $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=a$, where c is from $(x-1,x)$ (mean value theorem). As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=a$. Which is a contradictition since we have that  $\lim_{x \to \infty}f'(x)=+\infty$ .   2) I will also prove this by contradiction. We have that $\lim_{x \to \infty}f(x)=a$ This is equivalent to $\lim_{x \to \infty}f(x)-f(x-1)=0$. I use mean value theorem and get $\lim_{x \to \infty}(f(x)-f(x-1))=\lim_{x \to \infty}\frac{f(x)-f(x-1)}{x-(x-1)}=\lim_{x \to \infty}f'(c)=0$, where c is from $(x-1,x)$. As $x$ goes to infinity so does $c$ i.e. we have that $\lim_{x \to \infty}f'(c)=\lim_{c \to \infty}f'(c)=0$. Which is a contractition since we have that  $\lim_{x \to \infty}f'(x)=+\infty$ .
87861
If $f(x)$ is differentiable at x, I need to prove that $\lim_{h\to0}\frac{f(x+h)-f(x-h)}{2h}$ exist and is finite.  so if $f(x)$ is differentiable at a $x$, the difference quotient exist for this point, and also $f(x)$ must be continuous at x as well  so that mean that:  $\lim_{h\to0^+}\frac{f(x+h)-f(x)}{h} =  \lim_{h\to0^-}\frac{f(x+h)-f(x)}{h}$  and that: $\lim_{x\to x_0^+}=\lim_{x\to x_0^-}=f(x_0)$  I know I should probably use arithmetic limit laws to prove this but I can't see how what I figured out could help me. any help with that?  Thanks!
42594
I have a problem/proof that I can't quite solve.  Let $$f(x)=\begin{cases}x^2\quad &amp;x \in \mathbb{R} \setminus  \mathbb{Q}\\-x^2 \quad &amp;x \in \mathbb{Q}.\\ \end{cases}$$ Prove that $$\lim_{x\to n} f(x)$$ doesn't exist for $n$ other than $0$.  Maybe I can use the regular epsilon-delta proof? I tried that but wasn't sure of what assumptions I could make with the absolute values. I would appreciate any help.
13545
Problem:   If $f(x)$ is continous at $x=0$, and $\lim\limits_{x\to 0} \dfrac{f(ax)-f(x)}{x}=b$, $a, b$ are constants and $|a|&gt;1$, prove that $f'(0)$ exists and $f'(0)=\dfrac{b}{a-1}$.   This approach is definitely wrong:    \begin{align} b&amp;=\lim_{x\to 0} \frac{f(ax)-f(x)}{x}\\ &amp;=\lim_{x\to 0} \frac{f(ax)-f(0)-(f(x)-f(0))}{x}\\ &amp;=af'(0)-f'(0)\\ &amp;=(a-1)f'(0) \end{align}  I will show you a case why this approach is wrong:     \[f(x)= \begin{cases} 1,&amp;x\neq0\\ 0,&amp;x=0 \end{cases}\]   $\lim_{x\to0}\dfrac{f(3x)-f(x)}{x}=\lim_{x\to0} \dfrac{1-1}{x}=0$   but $\lim_{x\to0}\dfrac{f(3x)}{x}=\infty$,$\lim_{x\to0}\dfrac{f(x)}{x}=\infty$   Does anyone know how to prove it? Thanks in advance!
109480
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 $c$?