Find the Limit

Sorry .. is it 5050 ??

\hspace{-16}$Given $\bf{\lim_{x\rightarrow \infty}100\left\{\left[(x+1)\cdot(x+2)\cdot(x+3)\cdot......(x+100)\right]^{\frac{1}{100}}-x\right\}}$\\\\\\ Now Using $\bf{A.M\geq G.M}.$\\\\\\ $\bf{\frac{(x+1)+(x+2)+........+(x+100)}{100}\geq \left\{(x+1)\cdot(x+2)\cdot(x+3)......(x+100)\right\}^{\frac{1}{100}}}$\\\\\\ and equality hold when $\bf{(x+1)=(x+2)=.......=(x+100)}$\\\\\\ Now When $\bf{x\rightarrow \infty}$, each terms are equal.\\\\\\ Means $\bf{\lim_{x\rightarrow \infty}(x+1)=\lim_{x\rightarrow \infty}(x+2)=........=\lim_{x\rightarrow \infty}(x+100)}$\\\\\\ So when $\bf{x\rightarrow \infty},$ Then $\bf{A.M=G.M}$\\\\\\So $\bf{\lim_{x\rightarrow \infty}\frac{100x+\frac{100\cdot 101}{2}}{100}= \lim_{x\rightarrow \infty}\left[(x+1)\cdot(x+2)\cdot(x+3)......(x+100)\right]^{\frac{1}{100}}}$\\\\\\ So $\bf{\lim_{x\rightarrow \infty}\left\{(x+50.5)-x\right\} = \lim_{x\rightarrow \infty}\left\{\left[(x+1)\cdot(x+2)\cdot(x+3)\cdot......(x+100)\right]^{\frac{1}{100}}-x\right\}}$\\\\\\

\hspace{-16}$So $\bf{\lim_{x\rightarrow \infty}\left\{\left[(x+1)\cdot(x+2)\cdot(x+3)\cdot......(x+100)\right]^{\frac{1}{100}}-x\right\}=50.5}$\\\\\\ So $\bf{\lim_{x\rightarrow \infty}100\times \left\{\left[(x+1)\cdot(x+2)\cdot(x+3)\cdot......(x+100)\right]^{\frac{1}{100}}-x\right\}=100 \times 50.5 = 5050}$

hw did u find it??