# Not homework, just curious

1. Sep 20, 2005

### bomba923

*Let $f\left( x \right)$ be a twice-differentiable function for which
$$\; \mathop {\lim }\limits_{x \to 0^ - } f\left( x \right) = \infty$$

Then, is it true that
$$\mathop {\lim }\limits_{x \to 0^ - } f\,{''}\left( x \right) > 0\;?$$

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Or a little differently,

*Let $f\left( x \right)$ be a twice-differentiable function for which
$$\mathop {\lim }\limits_{x \to 0^ - } f\left( x \right) = \infty \;{\text{and }}\forall x < 0,f\,'\left( x \right) > 0$$

Then, is it true that
$$\mathop {\lim }\limits_{x \to 0^ - } f\,{''}\left( x \right) > 0\;?$$

Just curious|

Last edited: Sep 20, 2005
2. Sep 20, 2005

### VietDao29

You can prove it by contradiction.
So let f(x) be a function that is continuous, increasing in [-δ, 0), its first and second derivatives are also continuous in [-δ, 0), where δ > 0, so f(-δ) <> +∞, f'(-δ) <> +/-∞, f''(-δ) <> +/-∞.
So assume $$\lim_{x \rightarrow 0 ^ -} f''(x) <= 0$$
Now $$\lim_{x \rightarrow 0 ^ -} f''(x) = 0$$ means that there exists ε > 0, and 0 <= σ < ∞, such that, $\forall x \in (-\varepsilon,\ 0), \ f'(x) \leq \sigma$, where 0 <= σ < ∞
Since f(-ε) <> +/-∞, so f(0) <= f(-ε) + σε < + ∞.
Now $$\lim_{x \rightarrow 0 ^ -} f''(x) < 0$$ means that there exists ε > 0 such that, $\forall x_1, x_2 \in (-\varepsilon,\ 0) \mbox{ and } \ x_1 < x_2 \rightarrow \ f'(x_1) > f'(x_2)$
Let $$x_2 = \lim_{\beta \rightarrow 0 ^ -} 0 + \beta$$
Let α be the slope of the line that connects two points (x1, f(x1)), and (x2, f(x2)), so
f'(-ε) > f'(x1) > α > f'(x2) > 0 (f(x) is increasing), so α <> +/-∞.
(You can use: $$\forall \alpha , \ \beta | \alpha < \beta, \exists \gamma | \ \alpha < \gamma < \beta,\frac{f(\alpha) - f(\beta)}{\alpha - \beta} = f'(\gamma)$$ to prove that).
So f(x2) = f'(x1) + α(x2 - x1) <> + ∞, f(x2) does not tend to infinity.
(Q.E.D).
So, if:
$$\lim_{x \rightarrow 0 ^ -} f(x) = + \infty$$, then:
$$\lim_{x \rightarrow 0 ^ -} f''(x) > 0$$.
Note that it should be + ∞, (not - ∞).
Do the same, and you will get:
$$\lim_{x \rightarrow 0 ^ -} f(x) = - \infty$$, then:
$$\lim_{x \rightarrow 0 ^ -} f''(x) < 0$$.
To me the 0- is can be change to γ- for a more general case, where $\gamma \neq \pm \infty$. It's still true.
Viet Dao,

Last edited: Sep 21, 2005