Is the Limit of the Second Derivative Greater Than Zero?

[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\;?

----------------------------------------------
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|

 

[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&#039;&#039;(x) &lt;= 0
Now \lim_{x \rightarrow 0 ^ -} f&#039;&#039;(x) = 0 means that there exists ε > 0, and 0 <= σ < ∞, such that, \forall x \in (-\varepsilon,\ 0), \ f&#039;(x) \leq \sigma, where 0 <= σ < ∞
Since f(-ε) <> +/-∞, so f(0) <= f(-ε) + σε < + ∞.
Now \lim_{x \rightarrow 0 ^ -} f&#039;&#039;(x) &lt; 0 means that there exists ε > 0 such that, \forall x_1, x_2 \in (-\varepsilon,\ 0) \mbox{ and } \ x_1 &lt; x_2 \rightarrow \ f&#039;(x_1) &gt; f&#039;(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 &lt; \beta, \exists \gamma | \ \alpha &lt; \gamma &lt; \beta,\frac{f(\alpha) - f(\beta)}{\alpha - \beta} = f&#039;(\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&#039;&#039;(x) &gt; 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&#039;&#039;(x) &lt; 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,