Question

Evaluate ${lim}_{x\to \infty }(\sqrt[3]{3(x+1)(x+2)(x+3)}-x)$

Answer 1

$\underset{x\to \infty }{lim}\left(\right(2\left)\right(x+1\left)\right(x+2\left)\right(x+3)-x)$

Now, $\left(\right(x+1\left)\right(x+2\left)\right(x+3){)}^{\gamma 3}$

$=(x^3+6x^2+11x+6{)}^{\gamma 3}$

Now $\underset{x\to \infty }{lim}\left[\right(x^3+6x^2+11x+6{)}^{\gamma 3}-x]$

$\underset{x\to \infty }{lim}\left[\right(x^3{)}^{\gamma 3}{(1+\frac{6x^2}{x^3}+\frac{11x}{x^3}+\frac{6}{x^3})}^{\gamma 3}-x]$

$\underset{x\to \infty }{lim}[x{(6+\frac{6}{x}+\frac{11}{x^2}+\frac{6}{x^3})}^{\gamma 3}-x]$

Now $\underset{x\to \infty }{lim}\frac{6}{x}=0,\underset{x\to \infty }{lim}\frac{6}{x^3}-0$

$\underset{x\to \infty }{lim}\frac{11}{x^2}=0$

$\therefore \underset{x\to \infty }{lim}\left[x\right(1)-x]=0$

Hence the value of limit is $0$