Question

If the value of $\underset{x\to \infty }{limx}\left(3e-\left(\frac{e(x+2)}{x+1}\right)\right)$ equals $\frac{ae}{b}$ where $\frac{a}{b}$ is a rational in its lowest form, find the value of $\left(a^4+b^5\right)$.

Answer 1

$\underset{x\to \infty }{limx}\left(3e-\left(\frac{e(x+2)}{x+1}\right)\right)=3e-\frac{e}{1}=2e$

Hence, $a=2$ and $b=1$.

So, $a^4+b^5=2^4+1^5=17$