Proof verification needed for interesting advanced calculus problem. 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?
 A: 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.
