vertical asymptote of derivatives function 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 $$\lim\limits_{x \to 0^+} f'(x)=-\infty $$
 A: 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,
A: Notice that $$f''(x)>0$$
It means $f'(x)$ is increasing function, and then, the shape of function $f(x)$ is like this.(approximatlly)

so, you can notice that $$\lim_{x\rightarrow 0^+}f'(x)=-\infty$$
A: 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. 
A: 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
$$
\mathop {\lim }\limits_{x \to 0^ +  } f'(x) =  - \infty .
$$
A: 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 $x\to0$.
