Question

solve the problem in the diagram...

Answer 1

$\lim_{x \to \infty} { \sqrt[99]{(x+1)(x+2)(x+3)......(x+99)} - x }$
let y = 1/x  
The  limit as  x -> ∞ ,   y -> 0.


So the question now transformed into:
$\lim_{y \to 0} \frac{1}{y} {[ \sqrt[99]{(1+y)(1+2y)(1+3y)...(1+99y)} - 1 ]}\\\\F(y)=\sqrt[99]{(1+y)(1+2y)(1+3y)...(1+99y)}\\\\Ln\ F(y)=\frac{1}{99}[Ln(1+y)+Ln(1+2y)+.....+Ln(1+99y)]\\\\\lim_{y \to 0} Ln\ F(y)\\=\frac{1}{99}[\lim_{y \to 0}Ln(1+y)+\lim_{y \to 0}Ln(1+2y)+...\\\\. \ \ +\lim_{y \to 0}Ln \ (1+99y)]\\\\=\frac{1}{99}[ y+ 2y+3y+4y+.....99y ]\\\\=\frac{1}{99}*\frac{99*100}{2}y=50y\\\\we\ used\ \lim_{y \to 0}Ln(1+y)=y \\\\So F(y)= e^{50y}= 1+50 y [1+50y/2! +50^2y^2/3!+.....\infty]$

$so \lim_{y \to 0} \frac{1}{y} [F(y) -1]= \frac{1}{y}*50y*[1+y*(polynomial\ in\ y)]\\\\=50$