Evaluating the limit: $\lim_{x\rightarrow\infty}\frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e^n}-\frac{x(b^{n+2}e^n-a^{n+2}d^n)}{e^{2n}}$ Can someone please evaluate this limit for me, I have been breaking my head for the past 2-3 days..... Any help would be appreciated.
$$\lim_{x\rightarrow\infty}\left \{ \frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e^n}-\frac{x(b^{n+2}e^n-a^{n+2}d^n)}{e^{2n}} \right \}
$$
$a,b,d,e$ are constants; $e\neq 0 $.
$ (a,b,c,d)  \in  \mathbb{R}$
 A: I think that $n \in \mathbb{N}^*$
Lemma : Let $f$, $g$ and $h$ be positive function for sufficiently large $x$, if $f \sim g$ (ie $\lim_{x \to \infty} \frac f g=1)$ then $f+h \sim g+h$
(Sketch of proof) We estimate $\frac{f+h}{g+h}=1+\frac{f-g}{g+h}=\frac{g}{g+h}(\frac{f}{g}-1)$. This does not converges to $0$ if $\frac{g}{g+h}$ tends to some infinity, which is not possible as $g$, $h$ positive functions for sufficiently large $x$ as $g(x) \le g(x)+h(x)$ for such $x$
We obviously have $ \frac{(ax+b)^{n+2}}{(ex+d)^n} \sim \frac{(ax)^{n+2}}{(ex)^n}=\frac{a^{n+2}}{e^n} x^2$ and by our lemma the limit is equivalent (by $\sim$) to
$\lim_{x \to \infty}\frac{x(b^{n+2}e^n-a^{n+2}d^n)}{e^{2n}}$.
So the limit is of the form $\lim_{x \to \infty} \alpha x$, for $\alpha$ a constant, the rest should be easy.
A: Set $p=b^{n+2}e^n-a^{n+2}d^n$
\begin{align}
  &\lim_{x\to\infty}\left\{\frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e^n}-\frac {x(b^{n+2}e^n-a^{n+2}d^n)}{e^{2n}}\right\}=\\
  &\qquad\lim_{x\to\infty}\left\{\frac{(ax+b)^{n+2}}{(ex+d)^n}-\frac{a^{n+2}x^2}{e ^n}-\frac{px}{e^{2n}}\right\}=\\
  &\qquad\lim_{x\to\infty}\frac{e^{2n}(ax+b)^{n+2}-e^na^{n+2}x^2(ex+d)^n-px( ex+d)^n}{e^{2 n}(ex+d)^n}
\end{align}
now expand the numerator by the binomial theorem
\begin{align}
  &\sum_{k=0}^{n+2}\binom{n+2}{k}a^{k}b^{n+2-k}e^{2n}x^k-\sum_{k=0}^n\binom{n}{k}a ^{n+2}e^{n+k}d^{n-k}x^{k+2}-\sum_{k=0}^n\binom{n}{k}e^kd^{n-k}px^{k+1}=\\
  &\qquad=\sum_{k=0}^{n+2}\binom{n+2}{k}a^{k}b^{n+2-k}e^{2n}x^k-\sum_{k=2}^{n+2}\binom{n}{k-2}a^{n+2}e^{n+k-2}d^{n-k+2}x^k+{}\\
  &\qquad\qquad-\sum_{k=1}^{n+1}\binom{n}{k-1}e^{k-1}d^{n-k+1}px ^k=\\
  &\qquad=(a^{n+2}e^{2n}-a^{n+2}e^{2n})x^{n+2}+\sum_{k=2}^{n+1}\left[\binom{n+2}{k}a^{k}b ^{n+2-k}e^{2n}+\right.{}\\
  &\qquad\qquad\left.-\binom{n}{k-2}a^{n+2}e ^{n+k-2}d^{n-k+2}-\binom{n}{k-1}e^{k-1}d^{n-k+1}p\right]x^k+{}\\
  &\qquad\qquad+[(n+2)ab^{n+1}e^{2 n}-d^np]x+b^{n+2}e^{2n}
=\\
\end{align}
we see that the leading term cancels. The denominator has a non vanishing leading term $e^{3n}x^n,$  while in the numerator remains a term of degree $n+1,$ and if its coefficient does not vanish the limit goes to $\infty.$ The coefficient is given by
$$
(n+2)a^{n+1}be^{2n}+na^{n+2}e^{2n-1}d-e^np
$$
If this coefficient is null, numerator and denominator have the same degree, provided that the next coefficient in the numerator is not null, so the limit is a non vanishing constant, given by
$$
\frac{1}{e^{3n}}\left(\frac{(n+2)(n+1)}{2}a^nb^2e^{2n}-\frac{n(n-1)}{2}a^{n+2}d^2e^{2n-2}-nde^{n-1}p\right)
$$
If also this coefficient vanishes, then the numerator has degree less tha $n$ and the limit is $0.$
