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let $f:(0,\infty) \rightarrow \infty$ have the following properties: (I suppose $f$ continous)

a.) $\lim_{x \rightarrow \infty} \dfrac{f(x)}{x^k}=a ,a \in \mathbb R \bigcup \infty$

b.) $\lim_{x \rightarrow \infty} \dfrac{f(x+1)-f(x)}{x^{k-1}}=b , b \in \mathbb R \bigcup \infty$

c.) with $k \in \mathbb N_{>0}$ and different from $1$.

show that: $ \ b=ka$ , My approach:

$b=\lim_{x \rightarrow \infty} \dfrac{f(x+1)-f(x)}{x^{k-1}}=\lim_{x \rightarrow \infty} \dfrac{f(x+1)(x+1)^k}{x^{k-1}(x+1)^k}-\dfrac{f(x)x}{x^k}=\lim_{x \rightarrow \infty} \dfrac{f(x+1)(x+1)^k}{x^{k-1}(x+1)^k}-\lim_{x \rightarrow \infty}\dfrac{f(x)x}{x^k}=a \cdot \lim_{x \rightarrow \infty}\dfrac{(x+1)^k}{x^{k-1}}-x = \cdot a \lim_{x \rightarrow \infty}\dfrac{(x+1)^k-x^k}{x^{k-1}}$

and from here (using binomial expanson) we get : $b=a\cdot \lim_{x \rightarrow \infty} \binom{k}{1}=k$.

I'm unsure if I can subtract and recompose the limits like that. I know that if $f$ is continous ( wich i took as granted) we can split limits in this case but can we recompose them as i did? Some explanations/feedback on this would be very helpful,thanks.

-Also I suppose that $f$ is continous , would my reasoning apply if $f$ is not continous?

-Would the result of the problem hold if $f$ is not continous?

sigmatau
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  • I think the part a) should look like $\lim_{x \to \infty}\dfrac{f(x)}{x^{k}} = a$ and b) should be $\lim_{x \to \infty}\dfrac{f(x + 1) - f(x)}{x^{k - 1}} = b$. Please check and edit the post. – Paramanand Singh Nov 07 '13 at 16:37
  • After the edit question looks OK, but your manipulation are not valid according to rules of limits. The result looks intuitively obvious and perhaps you have followed your intuition. I am trying to figure out a rigorous proof which might need some extra conditions on $f$. – Paramanand Singh Nov 08 '13 at 09:58
  • Continuity is not required as shown by my solution. What is really needed is that the two limits $a$ and $b$ exist and $k$ be a positive integer. – Paramanand Singh Nov 08 '13 at 11:20
  • related to https://math.stackexchange.com/questions/192963 – Nosrati Sep 23 '18 at 07:27

1 Answers1

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After the edit of the question, we have

1) $\displaystyle \lim_{x \to \infty}\frac{f(x)}{x^{k}} = a$

2) $\displaystyle \lim_{x \to \infty}\frac{f(x + 1) - f(x)}{x^{k - 1}} = b$

From 1) we get that $f(x) = x^{k}(a + \rho(x))$ where $\rho(x)$ is an expression which tends to zero as $x \to \infty$. Similarly $f(x + 1) = (x + 1)^{k}(a + \rho(x + 1))$ where $\rho(x + 1) \to 0$ as $x \to \infty$.

Then

$\displaystyle \begin{aligned}b &= \lim_{x \to \infty}\frac{f(x + 1) - f(x)}{x^{k - 1}}\\ &= \lim_{x \to \infty}\frac{(x + 1)^{k}(a + \rho(x + 1)) - x^{k}(a + \rho(x))}{x^{k - 1}}\\ &= a\lim_{x \to \infty}\frac{(x + 1)^{k} - x^{k}}{x^{k - 1}} + \lim_{x \to \infty}\frac{\rho(x + 1)(x + 1)^{k} - \rho(x) x^{k}}{x^{k - 1}}\\ &= ak + \lim_{x \to \infty}\frac{\rho(x + 1)(x + 1)^{k} - \rho(x) x^{k}}{x^{k - 1}}\end{aligned}$

and this means that $$\lim_{x \to \infty}\frac{\rho(x + 1)(x + 1)^{k} - \rho(x) x^{k}}{x^{k - 1}}$$ exists. Now we can see that

$\displaystyle \begin{aligned}\lim_{x \to \infty}\frac{\rho(x + 1)(x + 1)^{k} - \rho(x) x^{k}}{x^{k - 1}} &= \lim_{x \to \infty}\frac{(\rho(x + 1) - \rho(x))x^{k} + \rho(x + 1)(kx^{k - 1} + \cdots)}{x^{k - 1}}\\ &= \lim_{x \to \infty}(\rho(x + 1) - \rho(x))x\end{aligned}$

If we can show that this above limit is zero then we are done. We are only given that it exists and is equal to $(b - ak)$. Let $\phi(x) = x\rho(x)$ then we can see that $\displaystyle \lim_{x \to \infty}(\rho(x + 1) - \rho(x))x = b - ak$

$\displaystyle \Rightarrow \lim_{x \to \infty}x\rho(x + 1) - x\rho(x) = b - ak$

$\displaystyle \Rightarrow \lim_{x \to \infty}(x + 1)\rho(x + 1) - x\rho(x) - \rho(x + 1) = b - ak$

$\displaystyle \Rightarrow \lim_{x \to \infty}(x + 1)\rho(x + 1) - x\rho(x) = b - ak$

$\displaystyle \Rightarrow \lim_{x \to \infty}\phi(x + 1) - \phi(x) = b - ak$

It can then be shown that $\lim_{x \to \infty}\phi(x)/x = b - ak$ or $\lim_{x \to \infty}\rho(x) = b - ak$ or $0 = b - ak$ and we are done.

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