Question

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

• A. 3
• B. 9
• C. 1
• D. none of these

Answer 1

Answer: (B)

9

$\underset{x\to \infty }{lim}{\left(\frac{3x^2+2x+1}{x^2+x+2}\right)}^{\frac{6x+1}{3x+2}}$
$=\underset{x\to \infty }{lim}{\left(\frac{3+2/x+1/x^2}{1+1/x+2/x^2}\right)}^{\frac{6+1/x}{3+2/x}}$
$=\underset{x\to \infty }{lim}{\left(\frac{3+2/x+1/x^2}{1+1/x+2/x^2}\right)}^{\underset{x\to \infty }{lim}\frac{6+1/x}{3+2/x}}$
$=3^2=9$
Hence, option 'B' is correct.