<s>[INST] 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
 [/INST] Hint:
Your method can be completed like this
$$HM \le GM \le AM \implies$$
$$\frac5{\frac1{x+1}+\frac1{x+2}+\frac1{x+3}+\frac1{x+4}+\frac1{x+5}} \le \sqrt[5]{(x+1)(x+2)(x+3)(x+4)(x+5)} \le x+3$$
Subtracting $x$ throughout :
$$\frac{15x^4+170x^3+675x^2+1096x+600}{5x^4+60x^3+255x^2+450x+274} \le \sqrt[5]{(x+1)(x+2)(x+3)(x+4)(x+5)} -x \le 3$$
Now take the limit as $x \to \infty$.
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