Evaluation of $\lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$ 
Evaluation of $\displaystyle \lim_{x\rightarrow \infty}\left\{\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right\}$

$\bf{My\; Try::}$ Here $(x+1)\;,(x+2)\;,(x+3)\;,(x+4)\;,(x+5)>0\;,$ when $x\rightarrow \infty$
So Using $\bf{A.M\geq G.M}\;,$ We get $$\frac{x+1+x+2+x+3+x+4+x+5}{5}\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$x+3\geq \left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}$$
So $$\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\leq 3$$
and equality hold when $x+1=x+2=x+3=x+4=x+5\;,$ Where $x\rightarrow \infty$
So $$\lim_{x\rightarrow 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3$$
Can we solve the above limit in that way, If not then how can we calculate it 
and also plz explain me where i have done wrong in above method
Thanks

Clearly $$(x + 1)(x + 2)(x + 3)(x + 4)(x + 5) = x^{5} + 15x^{4} + \cdots$$ and as we will show later we don't need to bother about the coefficients of $x^{3}, x^{2}, x, x^{0}$ in order to solve this problem.
Let $P(x)$ be a monic polynomial of degree $n$ and let the coefficient of $x^{n - 1}$ in $P(x)$ be $a$ so that $$P(x) = x^{n} + ax^{n - 1} + bx^{n - 2} + \cdots$$ We prove that $$\lim_{x \to \infty}\left\{\sqrt[n]{P(x)} - x\right\} = \frac{a}{n}$$ and thus for our current problem $n = 5, a = 15$ so that the desired limit is $a/n = 3$. Clearly if we set $g(x) = \sqrt[n]{P(x)}$ then we can see that $$\frac{g(x)}{x} = \sqrt[n]{\frac{P(x)}{x^{n}}} = \sqrt[n]{1 + \frac{a}{x} + \cdots } \to 1\text{ as }x \to \infty$$ and hence
\begin{align}
L &= \lim_{x \to \infty}\left\{\sqrt[n]{P(x)} - x\right\}\notag\\
&= \lim_{x \to \infty}\frac{P(x) - x^{n}}{g(x)^{n - 1} + xg(x)^{n - 2} + \dots + x^{n - 1}}\notag\\
&= \lim_{x \to \infty}\frac{ax^{n - 1} + bx^{n - 2} + \cdots}{g(x)^{n - 1} + xg(x)^{n - 2} + \dots + x^{n - 1}}\notag\\
&= \lim_{x \to \infty}\dfrac{a + \dfrac{b}{x} + \cdots}{\left(\dfrac{g(x)}{x}\right)^{n - 1} + \left(\dfrac{g(x)}{x}\right)^{n - 2} + \dots + \dfrac{g(x)}{x} + 1}\notag\\
&= \frac{a + 0 + \cdots}{1 + 1 + \cdots + n \text{ terms}}\notag\\
&= \frac{a}{n}
\end{align}
