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

\begin{align} &\lim_{x\to\infty}\left(\left((x+1)(x+2)(x+3)(x+4)(x+5)\right)^{\frac15}-x\right)\\ &=\lim_{x\to\infty}\cfrac{\left((1+\frac1x)(1+\frac2x)(1+\frac3x)(1+\frac4x)(1+\frac5x)\right)^{\frac15}-1}{\frac1x}\\ &=\lim_{h\to0}\cfrac{\left((1+h)(1+2h)(1+3h)(1+4h)(1+5h)\right)^{\frac15}-1}{h}\\ &=f'(0)\\ f(x)&=\left((1+x)(1+2x)(1+3x)(1+4x)(1+5x)\right)^{\frac15}\\ f(x)^5&=(1+x)(1+2x)(1+3x)(1+4x)(1+5x)\\ \left(f(x)^5\right)'&=5f(x)^4f'(x)\\&=f(x)\left(\frac1{1+x}+\frac2{1+2x}+\frac3{1+3x}+\frac4{1+4x}+\frac5{1+5x}\right)\\ f'(0)&=\frac{\left.\left(f(x)^5\right)'\right|_{x=0}}{5f(0)^4}\\ &=\frac{1+2+3+4+5}{{5f(0)^3}}=\frac{1+2+3+4+5}{5}=3 \end{align}

Let us shift the variable by $3$ and get

$$\lim_{x\to\infty}\sqrt[5]{(x-2)(x-1)x(x+1)(x+2)}-x+3=\lim_{x\to\infty}x\left(\sqrt[5]{1-\frac5{x^2}+\frac4{x^4}}-1\right)+3.$$

Then by L'Hospital,

$$\lim_{t\to0}\frac{\sqrt[5]{1-5t^2+4t^4}-1}t=\lim_{t\to0}\frac{-10t+16t^3}{5\sqrt[5]{1-5t^2+4t^4}}=0.$$

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 $x \to \infty$.