$\underset{x\to \mathrm{\infty }}{lim} {\left(\frac{3x^2+2x+1}{x^2+x+2}\right)}^{\frac{6x+1}{3x+2}}$is equal to 

Solution:

we observe that 

$\underset{x\to \mathrm{\infty }}{lim} \frac{3x^2+2x+1}{x^2+x+2}=3$ and $\underset{x\to \mathrm{\infty }}{lim} \frac{6x+1}{3x+2}=2$

$\therefore \underset{x\to \mathrm{\infty }}{lim} {\left(\frac{3x^2+2x+1}{x^2+x+2}\right)}^{\frac{6x+1}{3x+2}}=3^2=9$