Suppose $f$ is twice differentiable on $(a, \infty)$ and let $\lim_{x \to \infty}f(x) = 0$. Also assume that $f''(x)$ is bounded for all $x \in (a, \infty)$. Prove that $f'(x) \to 0$ as $x \to \infty$.   Note that if $f'(x)$ tends to a limit as $x \to \infty$ then this limit must be $0$ (because of the relation $f(x) - f(x/2) = (x/2)f'(c)$). What we need to establish now is that $f'(x)$ does tend to a limit. To show that we need to somehow use the fact that $f''(x)$ is bounded.   Let $b &gt; 0$ be any arbitrary but fixed number then we have relation $f(x + b) - f(x) = bf'(c)$ and this again implies that for all sufficiently large $x$ the derivative $f'$ takes a small value in the interval $(x, x + b)$. What we need to show that eventually all the values of $f'$ are small in such intervals.  Now we can take $b$ as small as we please, we can therefore find a value $N &gt; 0$  such that $f'$ takes very small value for at least one point in $(x, x + b)$ for all $x  &gt; N$. If $p, q$ are two points of this interval $(x, x + b)$ then $|f'(p) - f'(q)| = |(p - q)f''(\xi)| \leq b|f''(\xi)|$ and since $f''$ is bounded it follows that $f'(p) - f'(q)$ is small so that effectively all values of $f'$ in $(x, x + b)$ are small.  The above is a non-rigorous argument which I have not been able to make rigorous by using $\epsilon, \delta$ in proper manner. Maybe the approach used above can't be made rigorous and I am wrong path (or may be not!). Please help me in making above approach rigorous or suggest some alternative method.  Further there is another generalization available:     If $f(x) $ tends to a finite limit as $x \to \infty$ and $f^{(n + 1)}(x)$ is bounded then show that $f^{(n)}(x) \to 0$ as $x \to \infty$.   I have not tried to solve this general problem and any hint would be great.
26952
Let $f\colon\mathbb R\to\mathbb R$ be twice differentiable with $f(x)\to 0$ as $x\to\infty$ and $f''$ bounded. Show that $f'(x)\to0$ as $x\to\infty$. (This is inspired by  to a different question)
122170
Suppose $f(0) =0 $ and  $0&lt;f''(x)&lt;\infty (\forall$  $x&gt;0)$, then $\frac{f(x)}{x}$ strictly increases as $x$ increases.    I have shown that $f'(x)-\frac{f(x)}{x}  = \frac{1}{2}xf''(c)$, for some $c\in (0,x)$. How do I proceed from here?
42084
For the function $f(x) = \sqrt{x}, \hspace{2 mm} f: [0, \infty) \rightarrow \mathbb{R}$, I wish to find all points $a$ at which $f(x)$ is differentiable, from the definition $$ f'(x) = \lim_{x \rightarrow a}\frac{f(x)-f(a)}{x-a} $$  Clearly, for our function $f(x) = \sqrt{x}$, this gives us $$ f'(x) = \lim_{x\to a}\frac{\sqrt{x} - \sqrt{a}}{x-a} $$  So the question boils down to showing for what values of $a$ this limit exists. How would this be done? Intuitively, I know that the answer is that the limit exists for all $a \in [0, \infty)$ and is given by $\frac{1}{2 \sqrt{a}}$, but I'm unsure of how to prove this from the limit definition of differentiability.
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!
43481
PROBLEM. Let $f:[a,b]\to\mathbb R$ be a continuous function, such that $f'(x)\ge 0$, for all $x\in [a,b]\setminus A$, where $A\subset [a,b]$ is a countable set. Show that $f$ is increasing.  Attention. In this problem, we DO NOT assume that $f$ is differentiable in the whole $[a,b]$.  Notes. (1) If we assume that $f$ is differentiable in the whole interval, then we can easily show that $f'(x)\ge 0$, everywhere. For otherwise, if $f'(x_0)=c&lt;0$, for some $x_0\in [a,b]$, then by virtue of , $(c,0)\subset f'([a,b])$, and hence, $f'(x)&lt;0$, for uncountably many $x$'s.   (2) The conclusion of the problem does not hold if we replace the assumption $A$ is countable with $A$ is a set of measure zero. Take for example the , with a negative sign in front.  (3) If the hypothesis $f'(x)\ge 0$, is replaced by $f'(x)=0$, then the conclusion becomes f is constant.
51939
I started reading "The Mathematics of nonlinear Programming - AL peressini" recently.  Theorem : Suppose that $f(x) , f'(x) ,f''(x)$ exist on the closed interval $[a,b]$. If $x^* ,x $ are two different points of $[a,b]$ , then there exists a point $z$ strictly between $x^*$ and $x$ such that   $$f(x)=f(x^*)+f'(x^*)(x-x^*)+\frac{f''(z)}{2}(x-x^*)^2$$  I tried to prove this theorem using The Fundamental theorem of calculus. But I can not stop it at the second derivative . Do I have to use mean value theorem ? or is there a simple proof for this ?
26963
Question is :  For a differentiable function $f:\mathbb{R} \rightarrow \mathbb{R}$ what does $\lim_{x\rightarrow +\infty} f'(x)=1$ imply?  Options:    $f$ is bounded $f$ is increasing $f$ is unbounded $f'$ is bounded   I could not find counter examples but then, I strongly feel $\lim_{x\rightarrow +\infty} f'(x)=1$ would just imply that $f'(x)$ is bounded.  I do not have much idea why would other three are false.  I would be thankful if someone can suggest me some hints.  Thank You.
86705
Let $f:[0,1]\to\mathbb{R}$ be continuous with $f(0)=0$, $f(1)=1$ and $f$ is differentiable on $(0,1)$. Show that there are distinct $\xi,\eta\in(0,1)$ so that $f'(\xi)f'(\eta)=1$.  I think this requires mean value theorem. But this does not help since if apply the theorem to $f$ we can only get $f'(c)=1$ for some $c$.
62834
Let $f:\Bbb R\to (0,\infty)$ be a differentiable function such that for some constant $k$ we have, $$|f'(x)-f'(y)|\le k|x-y|$$   for all $x,y \in\Bbb R.$ Then prove that, $$(f'(x))^2&lt;2 kf(x).$$   Due to lack of arguments I could not proof this inequality. But rather I proved that is true for the particular function,   $$f(x) =\cos^2(x)+1\implies f'(x) = -\sin(2x).$$ And we readily have,$$|\sin(2x)-\sin(2y)| = \left|\int_{2x}^{2y}\cos t dt\right|\le 2|x-y|$$ as well   $$(f'(x))^2 = 4 \sin^2 x\cos^2x &lt; 4(\cos^2x +1) = 4f(x).$$  It is also true for the function $x\mapsto \sin^2x +1.$   From these example I don't see how to prove the general case. Any hint will be welcome.
59481
Let $f(x)$ be twice-differentiable and $f(a)=f(b)=0$ with $a&lt;b$  How to prove that  $$\max|f(x)| \le \frac 18(b-a)^2 \max|f''(x)|$$  (With $\max|f(x)|$ beeing maximum (or minimum) between a and b.)  How would I use Taylor-Expansion and Mean Value Theorem here?
45476
I'm trying to solve this question:     Let $f:[0,+\infty) \to \mathbb{R}$ be derivable and $\lim_{x \to +\infty} f'(x) = L$, then $\lim_{x\to \infty}\frac {f(x)}{x}=L$.   I'm trying to solve this question using l'Hôpital rule, but I couldn't use it, because I don't know if $\lim_{x\to +\infty} f(x)=+\infty$.  I need help.  Thanks a lot.
45139
Suppose for real numbers $a&lt;b$ one has a function with continuous derivative $f:[a,b] \to \mathbb{R}$ such that $f(a) = 0$ and there exists a real number $C$ with $$|f'(x)| \leq C|f(x)|$$ for all $x \in [a,b]$. Show that $f(x) = 0$ for all $x \in [a,b]$.  Well, since $f(a) = 0$, we have that $|f'(a)| \leq C|0|$, so $|f'(a)| = 0$. Since $f$ has a continuous derivative, we also know that $f$ is continuous. Since $f$ is continuous on a compact interval, $f$ obtains a maximum, say at $\xi \in [a,b]$. So, $|f'(x)| \leq C|f(\xi)|$. Since the derivative is bounded, we obtain that $f$ is Lipschitz, so $f$ is also uniformly continuous. Suppose to the contrary that $f(\xi) &gt; 0$? The above is pretty much everything I could figure out about $f$, so I'm not sure what to try next. This one is also from an old qual and possibly uses methods from beyond our course. I think maybe I should try to show that $|f'(x)| = 0$ for all $x$, but I don't know how.
71040
I am struggling with the proof of the following theorem:  Let $f(x)$ be uniformly continuous in $[a, \infty)$ s.t. the integral $\int_a^{\infty} f(x)dx$ converges. prove that $\lim_{x \to \infty} f(x) = 0$.  I came to the conclusion that it is enough to prove that $\lim_{x \to \infty} f(x)$ exists, and from there I have a proof that the limit is $0$.  I tried to use Cauchy's equivalent definition for the convergence of the improper integral + his definition for a regular limit at $x \to \infty$ with no success...  I will be happy to get hints as for how I should proceed, not proofs, since I really want to crack this one by myself, only I spent a pretty long time with no success.  Thank you :)
89465
I am struggling to construct a simple proof of the following.  Suppose $f(x)$ is differentiable for any real value $x$ and that $f'(x) \le 4$ for all real values of x. Prove there is at most one real value $c&gt;2$ such that $f(c) =c^2$.  I am completely lost on where to start on this proof and any help would be greatly appreciated!
102534
Suppose that $f:[a,b]\to\mathbb{R}$ be a twice-differentiable function, and that there exists $c\in(a,b)$ such that $f'(c)=\frac{f(b)-f(a)}{b-a}$. Show that if $f''(x)&gt;0$ for all $x\in[a,b]$ and $f''$ is strictly increasing on $[a,b]$, then $c&gt;\frac{a+b}{2}$.  Using Taylor's theorem, i solved the problem. But, i'd like to prove that using convexity. Give some comments or hints. Thank you!
106092
Suppose $f:\mathbb{R} \to \mathbb{R}$ is a differentiable function such that $\lim\limits_{x\to\infty}f'(x)=1$,then is it true necessarily true that $f(x)$ unbounded?   I think that it will always intersect $y=c$ for every $c\in \mathbb{R}$ and thus cannot be bounded!
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.