What Happens to the Limit of a Function as x Approaches Zero?

[anemone]

Evaluate $\displaystyle \lim_{{x}\to{0}}\dfrac{P(x^2)-P(x)}{P(x)-P(0)}$ if $P(x)$ is a strictly increasing and differentiable function with $P(0)=0$.

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[anemone]

Congratulations to the following members for their correct solutions::)

1. MarkFL
2. lfdahl
3. Francolino
4. greg1313

Solution from Francolino:

Spoiler

$$ \lim_{x \to 0} \frac {P(x^2)-P(x)}{P(x) - \underbrace{P(0)}_{=0}} = \lim_{x \to 0} \frac {P(x^2)}{P(x)} - 1 = \left ( \lim_{x \to 0} \frac {\overbrace{P(x^2)}^{\rightarrow 0}}{\underbrace{P(x)}_{\rightarrow 0}} \right ) - 1 \overset{(1)}{=} \underbrace{\left ( \lim_{x \to 0} \frac {2\cdot x\cdot \overbrace{P'(x^2)}^{>0}}{\underbrace{P'(x)}_{>0}} \right )}_{=0} - 1 \overset{(2)}{=} -1 $$

(1) $ P(x) $ is differentiable $ \Rightarrow P(x) $ is continuous $ \Rightarrow \lim_{x \to 0} P(x) = P(0) = 0 $.
(2) "$ P(x) $ is a strictly increasing" $ \Rightarrow P(x) > 0, \forall x $.

Solution from greg1313:

Spoiler

We have

\(\displaystyle \lim_{x\to0}\frac{P(x^2)-P(x)}{P(x)-P(0)}\)

This is an indeterminate form (0/0), therefore L'Hopital's rule may be applied:\(\displaystyle \lim_{x\to0}\frac{P(x^2)-P(x)}{P(x)-P(0)}=\lim_{x\to0}\frac{2xP'(x^2)-P'(x)}{P'(x)}\)

Since P(x) is strictly increasing, P'(x) is positive, hence

\(\displaystyle \lim_{x\to0}\frac{P(x^2)-P(x)}{P(x)-P(0)}=\lim_{x\to0}\frac{2xP'(x^2)-P'(x)}{P'(x)}=-1\)

 

[anemone]

Hi MHB,

I want to apologize because johng has brought to my attention that this problem might not work if $P'(0)=0$ and I have found I must agree with his point of view. The members whom I have announced as correctly answering the problem this past Monday will still be counted as correct. But, I have to say this problem is not a well thought out challenge problem and I will now post concerning the point about which johng has expressed his concern, and it appears that he is absolutely right.

The trouble with this argument is that it is valid only if $P'(0)\ne0$. Otherwise, L'Hôpital's Rule isn't applicable.

Even though we are given that $P$ is differentiable and strictly increasing, it does not follow that $P'(0)>0$. All we can say is that $P'(0)\ge0$ for all $x$. (A simple example is the function $P(x)=x^3$.) So the given hypothesis does not necessarily imply that $P'(0)>0$ as we mentioned in the argument.

When $P'(0)=0$, it's tempting to try to salvage the situation by applying again the L'Hôpital's Rule, if we do so, the limit \(\displaystyle \lim_{{x}\to{0}} \dfrac{P(x^2)}{P(x)}\) will equal \(\displaystyle \lim_{{x}\to{0}} \dfrac{2xP'(x^2)}{P'(x)}\) provided the latter exists. Now, we cannot hastily say that the limit of the denominator is $P'(0)$ because $P'$ is not given to be continuous at 0. But even if we assume this to be true, the difficulty still remains if $P'(0)=0$. In that case, we can differentiate the numerator and denominator once more and write the limit as \(\displaystyle \lim_{{x}\to{0}} \dfrac{2P'(x^2)+4x^2P''(x^2)}{P''(x)}\). But problems occur too, first, it is not given that $P$ is twice differentiable. And even if we assume that it is so, what guarantee do we have to say that $P''(0)$ is non zero? The trouble is that there do exist strictly increasing functions which have derivatives of all orders everywhere but all whose derivatives vanish at 0.

Hence, this is a wrongly set challenge question.:(

I want to thank johng again for taking the time to point this out to me by contacting me directly. Well done, johng for wanting to clarify regarding this issue and for making certain our solutions are as complete and error free as possible.