Lat $x &lt; y &lt; z$. Suppose $f : (x, z) \rightarrow \mathbb{R}$ is continuous and differentiable at every point of $(x, z)$ \ ${y}$, and that $f'$ has a limit at $y$. Prove that $f$ is differentiable at $y$ and that $f'(y) = \lim_{x \to y} f'(x)$.  How will we go about showing differentiability and the limit in this case? Firstly, I'm unsure what is meant by the fact that "$f'$ has a limit at $y$" and how is that relevant? Moreover, am I allow to use the mean-value theorem or does that require differentiability at y?
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.
37886
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...
1951
Question:     Let $f$ be differentiable on $[0,+\infty)$, such as$$\lim_{x\to+\infty}\left([f'(x)]^2+f^3(x)\right)=0$$show that $$\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$$   I think this problem is similar to  and (the  background)  For my problem I guess that if  $$\lim_{x\to+\infty}\left([f'(x)]^{2m}+f^{2n+1}(x)\right)=0,m,n\in N^{+},n,m&gt;1$$ then we have $\lim_{x\to\infty}f(x)=\lim_{x\to\infty}f'(x)=0$.
123647
$f$ is differentiable on $[a,b]$, $f&#39;(x) \leq A|f(x)|$ where $A$ is a non-negative constant.  If $f(a)=0$ show $f(x)=0, \forall x\in [a,b]$  I imagine the proof uses the Mean Value Theorem but I have not been able to get it to work.  I know $|f(x)|=|f&#39;(c)|(x-a)$ where $c \in [a,x]$, so $|f(x)| \leq A\  |f(c)|(x-a)$ where $c\leq x$ And I guess I could sort of iterate this to keep getting a smaller $c$ but I don't see why it must go all the way to zero.
79487
Let $f$ be differentiable on $\Bbb{R}$ and let $\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0$. Prove there exists $a\in \Bbb{R}$ such that $f'(a)=0$.  Attempt: If $f$ is constant we are done. Otherwise, $f$ is not constant. Suppose there is no such $a$. Then, in particular, $f$ has no extremum $\implies$ $f$ has no point where is goes from increasing to decreasing or from decreasing to increasing. $f$ is continuous since it is differentiable. Hence, since $\lim_\limits{x\to \infty}f(x)=\lim_\limits{x\to -\infty}f(x)=0$, $f$ must be constant. A contradiction. Therefore either $f$ is constant and $f'(x)=0$ everywhere, or $f$ has an extremum at $x=a$ and $f'(a)=0$.   I feel like my attempt is not formal enough, any correction?  I would appreciate your help.
26966
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.
119072
Question:   Let $f:[0,\infty)\rightarrow \mathbb{R}$ be differentiable and such that $\lim_{x\to\infty} f'(x) = L$. Then for every $c&gt;0$, $\lim_{x\to\infty} f(x+c)-f(x)=cL$ and $\lim_{x\to\infty} f(x)/x = L.$  What i think that may work: using mean value theorem for the first proof and l'hospital rule for the second proof. But i can't write this rigorously. How do i write those things in a rigorous way?
116800
I am trying to prove that if I have $f:\mathbb{R} \to \mathbb{R}$ satisfying $\forall x,y\in\mathbb{R},f(x+y) = f(x) + f(y)$. Which is assumed continuous at $0$, that $f$ is continuous on $\mathbb{R}$  I am fairly sure it is, for that property seems to be a property of polynomials, and we know polynomials are continuous where defined(all reals)  Any ideas for rigor?  $|x-a| \lt \delta \implies |f(x+a)-f(x)-f(a)| \lt \epsilon$  No idea where to go from here, this is my first time doing $\epsilon-\delta$ stuff.
111201
In this exercise one must show that $$f(x,y) = \begin{cases} 0 \text{     if } (x,y) = (0,0)\\ (x^2 + y^2) \sin\left(\frac{1}{x^2 + y^2}\right) \text{     else} \end{cases}$$ is differentiable for all $(x,y) \in \mathbb{R^2}$  In the lecture, we have proven that if all partial derivatives of $f$ exist and are continuous, then the derivative of $f(x,y)$ exists everywhere.  I start by calculating the partial derivatives of $f$ at $(0,0)$: $$\partial_xf(0,0) = \lim_{x\to0} \frac{f(x,0)-f(0,0)}{x} = \lim_{x\to0} \frac{x^2\sin({1 \over x^2})}{x} = 0$$ Similar for $\partial_y$: $$\partial_yf(0,0) = \lim_{y\to0} \frac{f(0,y)-f(0,0)}{x} = \lim_{y\to0} \frac{y^2\sin({1 \over y^2})}{y} = 0$$  We know that for all other values of $x$, $\partial_x f = 2x\sin\left(\frac{1}{x^2+y^2}\right)-\frac{2x\cos \left(\frac{1}{x^2+y^2}\right)}{x^2+y^2}$  I have tried to show that $\lim _ {x,y \to (0,0)} \partial_x f = 0$. (to show f is continuous) this however is problematic, since this limit doesn't exist. This implies that the partial derivates are not continuous at all points. Which makes this approach of proving that $f$ is differentiable useless.  How do I prove that $f$ is differentiable on all of $\mathbb{R}^2$.  I am open for all kind of suggestions. Thanks for your time.
103049
So. I have to show that this functiun is differentiable outside of $(0,0)$ and also to calculate  its differentiation. $$f(x,y)=xy\frac{x^2-y^2}{x^2+y^2}$$ when $(x,y)$ is not equal to $(0,0)$ and $$f(x,y)=0$$ when $(x,y)$ is equal to $(0,0)$. Also we have to show that the function is differentiable on $(0,0)$ too. $$$$ Firstly I wanted to see if the functiun is continuous e.g: $$\lim_{(x,y)\to (0,0)}{xy\frac{x^2-y^2}{x^2+y^2}}=0$$ so it is countinous.$$$$ Than I wanted to see if it is differentiable but I don't know how to continue, all I got is: $$\lim_{(x,y)\to (0,0)} \frac{f(x,y)-f(0,0)}{(x,y)-(0,0)}=\lim_{(x,y)\to (0,0)}\frac{xy\frac{x^2-y^2}{x^2+y^2}}{\sqrt{x^2+y^2}}$$ Can someone put me on the right track? And explain me what am I doing wrong? I don't know how to continue it..
80567
I came across the following challenging problem in my self-study:  Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be continuous. Then the set of points where $f$ is differentiable is a measurable set.  I am having trouble thinking of where to begin in proving this result, and wanted to see if anyone visiting had some suggestions on how to proceed.
1519
Let $f$ be differentiable at $x=a$.   Prove that if $x_n \to a^+$ and $y_n \to a^-$ then:  $$\lim_{n\to \infty} \frac{f(x_n)-f(y_n)}{x_n-y_n}=f'(a).$$  Every option that I think about seems to my very trivial, so I believe that I am doing something wrong.  Both numerator and denominator approach zero as $n\to\infty$ as the case of the formal definition of derivative, but it isn't guaranteed that the limits are equal (“$\frac{0}{0}$”).   Any direction?
79431
Let $f: \mathbb{R} \to \mathbb{R}$ be differentiable on $(a,b)$. Suppose $f' &gt; 0$ on $(a,b)$ except at a point $c \in (a,b)$ (that is, $f'(c) \leq 0$).  Is $f$ increasing on $(a,b)$? Must $f'(c)$ be zero, or can it be negative?  Clearly $f$ is increasing on $(a,c) \cup (c,b)$ but I'm not sure about how the value at $c$ compares with the values at other points. And I think $f'(c)$ must be zero: If $f'(c) &lt; 0$ then for small positive $h$ we have $\frac{f(c+h) - f(c)}{h}$ is also negative (by definition of the derivative as a limit of this ratio), so $f(c+h) - f(c) &lt; 0$. Since $f'$ is positive on $(c,c+h)$, the Mean Value Theorem implies that $f(c+h) - f(c) = f'(d)h$ for some $d \in (c, c+h)$, and $f'(d)h$ is a product of two positive numbers, hence positive. So $f(c+h) - f(c) &gt; 0$, a contradiction.
39174
Let $f:\mathbb R^2\to \mathbb R$ defined by $f(x,y)=xy$. Prove $f$ is differentiable.  According to the definition I have to show this $$\frac{\|f((x_0,y_0)+(h_1,h_2)) - f(x_0,y_0)-T(h_1\space  h_2)\|}{\|(h_1,h_2)\|}\to 0$$  But I got stuck in the proof.  \begin{align} &amp; \frac{\|f((x_0,y_0)+(h_1,h_2))-f(x_0,y_0)-T(h_1\space  h_2)\|}{\|(h_1,h_2\|} \\[10pt] = {} &amp; \frac{\|(x_0+h_1)(y_0+h_2)-x_0y_0-(x,y)(h_1\space  h_2)\|}{\sqrt{h_1^2+h_2^2}} \\[10pt] \le {} &amp;\frac{\|x_0h_2+y_0h_1+h_1h_2\|}{\sqrt{h_1^2+h_2^2}}+\frac{\|-(x,y)(h_1\space h_2)\|}{\sqrt{h_1^2+h_2^2}} \end{align} and I think I the first term after the inequality goes to zero, but what happen to the second term?  How can it goes to zero too?
87979
Claim: if $f$ is differentiable at $x = x_0$ then $f$ is continuous at $x = x_0$.   Please, see if I made some mistake in the proof below. I mention some theorems in the proof:   The condition to $f(x)$ be continuous at $x=x_0$ is $\lim\limits_{x\to x_0} f(x)=f(x_0)$.  (1) If $f(x)$ is differentiable at $x-x_0$, then $f'(x)=\lim\limits_{x\to x_0} \dfrac{f(x)-f(x_0)}{x-x_0}$ exists and the function is defined at $x=x_0$.   (2) Therefore, by the Limit Linearity Theorem, $\lim\limits_{x\to x_0} f(x)$ exists and we'll show it is equals $f(x_0)$.   (3) We'll do this by the Precise Limit Definion: given $ \epsilon&gt;0, \exists\delta|0&lt;|x-x_0|&lt;\delta$, then $0&lt;|f(x)-f(x_0)|&lt;\epsilon$. As this limit exists by (2), we can make $f(x)$ as close to  $f(x_0)$ as one wishes, therefore $\lim\limits_{x\to x_0} f(x)=f(x_0)$, what satisfies the condition for $f(x)$ be differentiable at  $x=x_0$. The end.
122767
Suppose that $\mathcal f$ is continuous on a closed interval and $\mathcal v$ is a point in its interior. $\mathcal f$ is differentiable when $\mathcal x\not=v$, and suppose that $\,\displaystyle{\lim_{x\rightarrow v}f'(x)}\,$ exists.  Show that $\mathcal f$ is differentiable at $\mathcal v$.  Attempt at a solution  I want to show that   $$ \lim_{h\rightarrow 0}\frac {f(v+h)-f(v)}{h}$$   exists. Based on the fact that   $$ \lim_{x\rightarrow v}f'(x)=\lim_{x\rightarrow v}\ \lim_{h\rightarrow 0}\frac {f(x+h)-f(x)}{h}$$  exists. Using the fact that $\mathcal f$ is continuous, I think if somehow I could interchange the limits $$\mathcal {\lim_{x\rightarrow v}\ \lim_{h\rightarrow 0}\frac {f(x+h)-f(x)}{h}}={\lim_{h\rightarrow 0}\ \lim_{x\rightarrow v}\frac {f(x+h)-f(x)}{h}}=\lim_{h\rightarrow 0}\frac {f(v+h)-f(v)}{h}$$ then the result comes out. But I don't think that I could do that, based on continuity. Other than that, I am not sure how to approach this.
52054
Consider the function $ f:\mathbb{R^2} \rightarrow \mathbb{R}$ given by   $ f(x,y) = \left\{   \begin{array}{l l}     \frac{xy}{x^2 + y^2} &amp; \quad \text{if (x,y) $\neq$ (0,0)}\\     0 &amp; \quad \text{if (x,y) = (0,0)}   \end{array} \right.$  Show that f is continuous along every horizontal and every vertical line (i.e. for every $x_0, y_0 \epsilon \mathbb{R}$, the functions $g,h:\mathbb{R}\rightarrow \mathbb{R}$ given by $g(t) = f(x_0,t)$ and $h(t)=f(t,y_0)$ are continuous).  I know that this function is not continuous at $(0,0)$, but the horizontal/vertical line issue is proving difficult to work with. Any help/hints appreciated!