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

Let $t=x+3$, then
$$
((x+1)(x+2)(x+3)(x+4)(x+5))^{\frac{1}{5}}-x=(t^5-5t^3+4t)^{\frac{1}{5}}-(t-3).
$$
Using $x^5-a^5=(x-a)(x^4+ax^3+a^2x^2+a^3x+a^4)$,
\begin{align}
&(t^5-5t^3+4t)^{\frac{1}{5}}-(t-3)\\
&=\frac{t^5-5t^3+4t-(t-3)^5}{(t^5-5t^3+4t)^{\frac{4}{5}}+(t^5-5t^3+4t)^{\frac{3}{5}}(t-3)+(t^5-5t^3+4t)^{\frac{2}{5}}(t-3)^2+(t^5-5t^3+4t)^{\frac{1}{5}}(t-3)^3+(t-3)^4}.
\end{align}
It looks complicated, and really it is. But we can compute limit comparing coefficients of $t^4$ with numerator and denominator. By binomial theorem, coeffecient of $t^5$ is $0$ and that of $t^4$ is $15$. If we divide numerator and denominator by $t^4$ and sends $t$ to $\infty$, numerator remains $15$. How about denominator? Denominator divided by $t^4$ is
$$
(1-5t^{-2}+5t^{-4})^{\frac{4}{5}}+(1-5t^{-2}+5t^{-4})^{\frac{3}{5}}(1-3t^{-1})+(1-5t^{-2}+5t^{-4})^{\frac{2}{5}}(1-3t^{-1})^2+(1-5t^{-2}+5t^{-4})^{\frac{1}{5}}(1-3t^{-1})^3+(1-3t^{-1})^4
$$
and goes to $1+1\cdot 1 + 1\cdot 1 + 1\cdot 1 +1=5$ as $t\to \infty$. Therefore,
$$
\lim_{x\to 0}\left[\left[(x+1)(x+2)(x+3)(x+4)(x+5)\right]^{\frac{1}{5}}-x\right]=3.
$$
