Question

Evaluate $\underset{x\to \infty }{lim}{\left(\frac{x^2+5x+3}{x^2+x+2}\right)}^x=e^a$. Find $a$

Answer 1

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

Let $y=\frac{4x+1}{x^2+x+2}=\frac{\frac{4}{x}+\frac{1}{x^2}}{1+\frac{1}{x}+\frac{2}{x^2}}\to 0$ as $x\to \infty$

Also

$xy=\frac{4x^2+x}{x^2+x+2}=\frac{4+1/x}{1+\frac{1}{x}+\frac{1}{x^2}}\to 4$ as $x\to \infty$

$\therefore lim=\underset{y\to 0}{lim}{(1+y)}^x$

$=\underset{y\to 0}{lim}{\left[{(1+y)}^{1/y}\right]}^{xy}=e^{xy}=e^4$ as $xy\to 4$..........As $\underset{y\to 0}{lim}\left[{(1+y)}^{1/y}\right]=e$

$\therefore$ $a=4$