Question

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

• A. $0$
• B. $\infty$
• C. $\frac{1}{2}$
• D. None of these

Answer 1

Answer: (C)

$\frac{1}{2}$

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

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

$=1/2$