Calculate the limit of function I was trying to find it for some time, but couldn't, so please help me.
$ \lim_{x\to\infty} ( \sqrt[100]{(x + 3*1)(x + 3*2)...(x +3*100)} - x)$
 A: Set $1/x=h$
$$\lim_{h\to0^+}\dfrac{\sqrt[100]{(1+3h)(1+6h)\cdots(1+300h)}-1}h$$
Now $(1+3h)(1+6h)=1+(3+6)h+O(h^2)$
$(1+3h)(1+6h)(1+9h)=(1+9h+O(h^2))(1+9h)=1+(3+6+9)h+O(h^2)$
Similarly, $(1+3h)(1+6h)\cdots(1+300h)=1+(3+6+9+\cdots+300)h+h^2=1+\dfrac{(3+300)100h}2+O(h^2)$
$\sqrt[100]{(1+3h)(1+6h)\cdots(1+300h)}=\left(1+\dfrac{(3+300)100h}2+O(h^2)\right)^{1/100}=1+\dfrac{303}2h+O(h^2)$
A: \begin{gather*}
\lim _{x\leadsto \infty }(\sqrt[100]{}( x+3\times 1)( x+3\times 2) ...( x+3\times 100) \ -\ x\\
Taking\ x\ common\ \ from\ all\ brackets\ ,\ \\
\lim _{x\leadsto \infty }( x\sqrt[100]{}( 1+3\times 1/x)( 1+3\times 2/x) ...( 1+3\times 100/x) \ -\ x\\
Apply\ the\ binomial\ series\ expansion\ because\ \frac{1}{x} \leadsto 0\\
\lim _{x\leadsto \infty }( x( 1+\frac{1}{100}\left(\frac{3\times 1}{x} +\frac{3\times 2}{x} +....\frac{3\times 100}{x}\right) \ -\ x\\
\lim _{x\leadsto \infty }\frac{1}{100}\left(\frac{3\times 1}{1} +\frac{3\times 2}{1} +....\frac{3\times 100}{1}\right) \ \ =\ \frac{3}{200}( 100\ \times 101) \ =\frac{303}{2} \ \\
\\
\\
\\
\end{gather*}
