If $p(x)=ax^n+b_1x^{n-1}+\cdots$ and $g(x)=ax^n+b_2x^{n-1}+\cdots$, then $\lim_{x\to\infty}{p(x)^{1/n}-g(x)^{1/n}}=\frac{b_1-b_2}{n\cdot a^{1/n}}$ 
Suppose $p(x)=ax^n + b_1x^{n-1}+\cdots~$ and $g(x)=ax^n + b_2x^{n-1}+\cdots~~$ (basically only the leading coefficients are same).
I am required to find/proof: $$\lim_{x \to \infty}{p(x)^{1/n}-g(x)^{1/n}}=\frac{b_1 - b_2}{n \cdot a^{1-(1/n)}}$$

I was only able to find it for $p(x)$ and $g(x)$ being quadratic ($n=2$), and just rationalizing the expression. I am not able to find for this though. Any help is appreciated.
I am also unaware if LHS even equals to RHS or if you can even generalise this. Our teacher was using the result for $n=3$, $n=4$, and so on. So I thought there is a generalized result for two polynomials with $n$ degree.
 A: Hint. Assuming that $a>0$, then as $x\to +\infty$,
$$(ax^n + bx^{n-1}+o(x^{n-1}))^{1/n}=a^{1/n}x\left(1+\frac{b}{ax}+o(1/x)\right)^{1/n}.$$
Now use the binomial approximation: $(1+t)^r=1+rt+o(t)$ as $t\to 0$.
Notice that the final result should be $\frac{b_1 - b_2}{n \cdot a^{1-1/n}}$. For $n>2$, it differs from yours!!
A: Let's write
$$
p(x)=ax^n+bx^{n-1}+p_0(x)
\qquad
g(x)=ax^n+cx^{n-1}+g_0(x)
$$
where $p_0$ and $g_0$ have degree $<n-1$. Now a trick is to replace $x=1/t$: we get
$$
p(1/t)=\frac{a+bt+t^np_0(1/t)}{t^n}=\frac{a+bt+t^2p_1(t)}{t^n}
$$
where $p_1$ is some polynomial.
Similarly
$$
g(1/t)=\frac{a+ct+t^2g_1(t)}{t^n}
$$
and your task has become to compute
$$
\lim_{t\to0^+}\frac{\sqrt[n]{a+bt+t^2p_1(t)}-\sqrt[n]{a+bt+t^2g_1(t)}}{t}
$$
which the derivative at $0$ of
$$
f(t)=\sqrt[n]{a+bt+t^2p_1(t)}-\sqrt[n]{a+bt+t^2g_1(t)}
$$
Or, with a standard Taylor expansion,
$$
\sqrt[n]{a+bt+t^2p_1(t)}=\sqrt[n]{a}\Bigl(1+\frac{b}{a}t+o(t)\Bigr)=
\sqrt[n]{a}\Bigl(1+\frac{b}{na}t+o(t)\Bigr)
$$
and so your limit is easily computed as
$$
\sqrt[n]{a}\Bigl(\frac{b}{na}-\frac{c}{na}\Bigr)=\sqrt[n]{a}\frac{b-c}{na}
$$
