Question

If $f\left(x\right)={\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x$ ,then $\underset{x\to \infty }{lim}f\left(x\right)$ is

• A. $e^4$
• B. $e^3$
• C. $e^2$
• D. $2^4$

Answer 1

The correct option is A $e^4$

For $f\left(x\right)={\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x$

$\Rightarrow \frac{x^2+5x+3}{x^2+x+2}=1+\frac{4x+1}{x^2+x+2}$

$\Rightarrow x=\frac{x^2+x+2}{4x+1}\times \frac{4+1/x}{1+1/x+1/x^2}$

Let $t=\frac{4x+1}{x^2+x+2}$
$\Rightarrow \underset{x\to \infty }{lim}f\left(x\right)=\underset{t\to 0}{lim}{\left({(1+t)}^{1/t}\right)}^{\underset{x\to 0}{lim}\frac{4+1/x}{1+1/x+1/x^2}}=e^4$