Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $$\lim\limits_{x \to 0^+} f'(x)=-\infty $$

share|improve this question
add comment

5 Answers

up vote 1 down vote accepted

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,

share|improve this answer
    
thank you I need this example but I have question.how about f'' >0 and f(x)→+∞ as x→0 . I guess f'(x)→-∞ as x→0 –  Babak Miraftab Jul 20 '11 at 15:23
    
...ask it? =) What's your question –  Patrick Da Silva Jul 20 '11 at 15:25
    
@Patrick: the OP edited the question, you may want to re-visit this answer. –  Willie Wong Jul 20 '11 at 18:25
    
I didn't remove my old answer because I thought it was helpful at first. I edited my answer to add the answer to the new question. –  Patrick Da Silva Jul 20 '11 at 20:32
    
Saw other people's answers... and this makes me realize I just love sequences, don't know why. Even though they make things longer sometimes. –  Patrick Da Silva Jul 20 '11 at 20:34
add comment

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 $$\lim_{x\rightarrow 0^+}f'(x)=-\infty$$

share|improve this answer
1  
Something is missing here... maybe rigor? –  The Chaz 2.0 Jul 20 '11 at 16:52
1  
Intuition is important. Thanks for the picture. +1 –  Patrick Da Silva Jul 20 '11 at 20:47
add comment

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.

share|improve this answer
    
What if the limit of $f'$ didn't converge to $\pm \infty$? You supposed it did, nothing shows it does.. =) (just messing with you haha) –  Patrick Da Silva Jul 20 '11 at 20:48
    
@Pat No, the above proof only assumes the existence of $\rm\:lim \ f'\:.$ –  Bill Dubuque Jul 20 '11 at 21:28
    
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. Heehee –  Patrick Da Silva Jul 20 '11 at 21:42
add comment

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 $$ \mathop {\lim }\limits_{x \to 0^ + } f'(x) = - \infty . $$

share|improve this answer
1  
Why are you posting a duplicate answer to a duplicate question? –  Alex B. Jul 20 '11 at 16:40
    
I will delete one of the answers later on. –  Shai Covo Jul 20 '11 at 16:44
    
Did it for you. ('cause I merged the two questions) –  Willie Wong Jul 20 '11 at 16:54
    
there are a problem in this solution!you can't solve limit in f(1) - f(x) \geq f'(x)(1-x) –  Babak Miraftab Jul 22 '11 at 11:20
    
@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 $x \to 0^+$. –  Shai Covo Jul 22 '11 at 11:33
show 1 more comment

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 $x\to0$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.