Question

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

Answer 1

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

Put $x\ln \left(\frac{x+2}{x+1}\right)-1=M$
$\therefore M\to 0$ as $x\to \infty$

$\Rightarrow L=-e\cdot \underset{M\to 0}{lim}x\left(\frac{e^M-1}{M}\right)M$
$\Rightarrow L=-e\cdot \underset{x\to \infty }{lim}x(x\ln \left(\frac{x+2}{x+1}\right)-1)(\because \underset{x\to 0}{lim}\frac{e^x-1}{x}=1)$

Put $x=\frac{1}{t}$
$\therefore L=-e\cdot \underset{t\to 0}{lim}\frac{\ln \left(\frac{1+2t}{1+t}\right)-t}{t^2}$

$\Rightarrow L=-e\cdot \underset{t\to 0}{lim}\frac{\ln (1+2t)-\ln (1+t)-t}{t^2}\left(\frac{0}{0}\text{form}\right)$
Using L'Hospitals rule, we get
$L=-e\cdot \underset{t\to 0}{lim}\frac{\frac{2}{1+2t}-\frac{1}{1+t}-1}{2t}\left(\frac{0}{0}\text{form}\right)$
Again using L'Hospitals rule, we get
$L=-e\cdot \underset{t\to 0}{lim}\frac{\frac{-4}{(1+2t{)}^2}+\frac{1}{(1+t{)}^2}}{2}$
$\Rightarrow L=\frac{3e}{2}=\frac{ae}{b}$
$\Rightarrow a=3$ and $b=2$
$\therefore a^4+b^5=81+32=113$