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Let $f$ be function that has derivatives of order $2$. Furthermore, $\lim\limits_{x \to 0^+} f(x)=+\infty $ and $f''(x)>0$. prove that $\lim\limits_{x \to 0^+} f'(x)=-\infty $

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The original question was : If $f$ has derivatives of order $2$ and $\lim_{x \to 0} f(x) = +\infty$, does that mean \lim_{x \to 0} f'(x) = -\infty and this is no.

Here : I'll suppose $f : \mathbb R / \{0\} \to \mathbb R$ and $f \to +\infty$ as $x \to 0^+$. Take $f(x) = \frac 1x + \sin \left( \frac 1{x^2} \right)$. You obtain f'(x) = \frac{-1}{x^2} -\frac{2 \cos \left( \frac 1{x^2} \right) }{x^{3}} = \frac{-x-\cos \left( \frac 1{x^2} \right)}{x^3} You can take subsequences such that the derivative goes to $-\infty$ as well as $+\infty$ (by letting $x_n$ such that $x_n \to 0$ and $\cos(\frac 1{x^2}) = \pm 1$) so that you don't have an asymptote for f', but rather some horrible behavior (oscillation between $-\infty$ and $+\infty$). Clearly this means the derivative does not necessarily converge to $+\infty$. But if it converges, it does indeed go to $+\infty$, because it can clearly not be bounded or go to $-\infty$ if $f$ goes to $+\infty$...

EDIT : Now for the current question, the answer is yes. Since f''(x) > 0, f'(x) is increasing, thus it suffices to find a subsequence of f' which goes to $-\infty$ as $x \to 0$. Consider the interval $(0,1)$. Define $x_1 = 1$ and choose $x_2$ in this interval so that $ f(x_2) - f(x_1) > 1. $ which is possible because $f$ goes to infinity. Suppose $x_n$ has been defined and choose $x_{n+1}$ in the interval $(0,x_n)$ such that $ f(x_{n+1}) - f(x_n) > n . $ By Taylor's theorem, there exists $c_n \in (x_{n+1},x_n)$ such that -n(x_n-x_{n+1}) > -n > f(x_n)- f(x_{n+1}) = f'(c_n)(x_n-x_{n+1}) which implies -n > f'(c_n) \frac{x_n - x_{n+1}}{x_n-x_{n+1}} = f'(x_n) and $c_n \to 0$ as $n \to \infty$. This gives you \forall n \in \mathbb N, \quad \exists c_n \text{ s.t. } \quad \forall 0 < x \le c_n, \quad f'(c_n) < -n. This means f'(x) \to -\infty.

Hope that helps,

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    Saw other people's a$n$swers... and this makes me realize I just love sequences, don't know why. Even though they make things longer sometimes.2011-07-20
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Notice that $f''(x)>0$ It means $f'(x)$ is increasing function, and then, the shape of function $f(x)$ is like this.(approximatlly)

enter image description here

so, you can notice that $\lim_{x\rightarrow 0^+}f'(x)=-\infty$

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    Intuition is important. Thanks for the picture. +$1$2011-07-20
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Note if $\rm\displaystyle\ \!\!\!\!\!\lim_{\quad x \to\: 0^+} f\:\:'(x)\ $ exists then applying L'Hospital's rule we deduce

$\rm 0\ =\ \!\!\!\!\!\lim_{\quad x \to\: 0^+}\:\dfrac{x}{ f(x)}\: =\ \!\!\!\!\!\lim_{\quad x \to\: 0^+} \dfrac{1}{f\:'(x)} $

hence $\rm\displaystyle \!\!\!\lim_{\quad x \to\: 0^+} f\:'(x)\ =\: \pm\:\infty,\ $ necessarily $\rm\:-\infty\:$ since $\rm\ f\:'' > 0\ \Rightarrow\ f\:'\:$ increasing.

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    Oh, I didn't notice your answer was not trying to answer but was assuming the limit existed. Sorry. I can be a loudmouth sometimes. Heehee2011-07-20
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By the mean-value theorem, for any $0< x <1$, f(1) - f(x) = f'(\xi)(1-x), for some $\xi \in (x,1)$. Since f'' > 0, f' is increasing. So from $\xi > x$, it follows that f'(\xi) \geq f'(x). Hence f(1) - f(x) \geq f'(x)(1-x). Now note that, by the assumption on $f$, $ \mathop {\lim }\limits_{x \to 0^ + } (f(1) - f(x)) = - \infty , $ and further note that $ \mathop {\lim }\limits_{x \to 0^ + } (1 - x) = 1 , $ to conclude from f(1) - f(x) \geq f'(x)(1-x) that \mathop {\lim }\limits_{x \to 0^ + } f'(x) = - \infty .

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    @babgen: There isn't a problem. Note that $f'(x) \le \frac{{f(1) - f(x)}}{{1 - x}} \to - \infty $ as $x \to 0^+$ to conclude that also $f'(x) \to - \infty$ as $x \to 0^+$.2011-07-22
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Here's how you can make ks0830's idea rigorous; I'll suppose $f$ is defined on the interval $(0,1]$:

Since $f$ is unbounded, f' must also be unbounded (otherwise $f$ would be Lipschitz continuous and hence bounded). Moreover, f''>0 implies that f' is increasing, so you conclude that f'(x) \to -\infty as $x\to0$.