Question

$\underset{x\to \infty }{\lim }{\left[\frac{x^2+x+3}{x^2-x+2}\right]}^X$ is equal to

• A. $\infty$
• B. $e$
• C. $e^4$
• D. $e^2$

Answer 1

Answer: (D)

Given, $\underset{x\to \infty }{\lim }{\left[\frac{x^2+x+3}{x^2-x+2}\right]}^x$
$=\underset{x\to \infty }{\lim }{\left[1-1+\frac{x^2+x+3}{x^2-x+2}\right]}^x$
$=\underset{x\to \infty }{\lim }{\left[1+\frac{x^2+x+3-x^2+x-2}{x^2-x+2}\right]}^x$
$=\underset{x\to \infty }{\lim }{\left[1+\frac{2x+1}{x^2-x+2}\right]}^x$
$=e^{\underset{x\to \infty }{\lim }\frac{2x+1}{x^2-x+2}\cdot (x)}$
$\left[\therefore \underset{x\to \infty }{\lim }[1+f(x){]}^{g(x)}=e^{\underset{x\to \infty }{\lim }f(x)\cdot g(x)}\right]$
$=e^{\underset{x\to \infty }{\lim }\frac{2x^2+x}{x^2-x+2}}=e^2$