Question

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

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

Answer 1

The correct option is D $e^2$

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

Divide by $x^2$ inside $f\left(x\right)$ for both numerator and denominator.

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

$=\frac{1+\frac{1}{\infty }+\frac{3}{{\infty }^2}}{1-\frac{1}{\infty }+\frac{2}{{\infty }^2}}=1$

$\underset{x\to \infty }{lim}g\left(x\right)=$$\infty$

This limit is of the form $1^{\infty }$ if $\underset{x\to \infty }{lim}{f\left(x\right)}^{g\left(x\right)}=1^{\infty }$

Then we can rewrite the limit as $\underset{x\to \infty }{lim}{e}^{\left(f\right(x)-1)g\left(x\right)}$

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

Now divide by $x^2$ inside $

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

Applying limits

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

$\Rightarrow e^2$