Infinite limit of sum with ($0\times \infty$) form and $1^{\infty}$ form limits and series to be needed

Question

I needed help with Part (A) without using L'Hopital's because its getting too lengthy.Can someone help me obtain solution with series without using L Hospitals rule

I'm trying something out with series

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Answer 1

Make the change of variables $y=\frac1x$ and this becomes $$\lim_{x\to\infty}\frac xe-x\left(\frac x{x+1}\right)^x=\lim_{y\to 0^+}\frac{\frac1e-(y+1)^{-1/y}}y$$

Now, of course this is an indeterminate form $\left[\frac 00\right]$ and the fraction is OK for l'Hopital. Therefore we can inspect $\frac{f'}{g'}$ \begin{align}&\lim_{y\to0^+}-\left(-\frac1{y(y+1)}+\frac{\ln(y+1)}{y^2}\right)(y+1)^{-1/y}&\\&=\lim_{y\to0^+}\frac{y-(y+1)\ln(y+1)}{y^2}\cdot\frac{(y+1)^{-1/y}}{y+1}\end{align}

$\lim_{y\to0^+}\frac{y-(y+1)\ln(y+1)}{y^2}$ is an indeterminate form $\left[\frac 00\right]$, therefore we can evaluate $\lim_{y\to 0^+}\frac{1-\ln(y+1)-1}{2y}=-\frac12$ and, by l'Hopital, $$\lim_{y\to 0^+}\frac{y-(y+1)\ln(y+1)}{y^2}=-\frac12.$$ Therefore $$\lim_{y\to 0^+}\frac{y-(y+1)\ln(y+1)}{y^2}\cdot \frac{(y+1)^{-1/y}}{y+1}=\left[-\frac12\cdot \frac{e^{-1}}{1}\right]=-\frac1{2e}$$

Which means that by l'Hopital $$\lim_{y\to 0^+}-\frac{(y+1)^{-1/y}-\frac1e}{y}=-\frac1{2e}$$

Answer 2

For $x>1$ we have $$x\ln \frac {x+1}{x}=x\ln \left(1+\frac {1}{x}\right)=x(x^{-1}-x^{-2}/2+x^{-3}/3-...)=1-x^{-1}/2+x^{-2}/3-...=$$ $$=1-[x^{-1}/2][1+F(x)]$$ where $F(x)\to 0$ as $x\to \infty.$

So we have$$\left(\frac {x+1}{x}\right)^x=\exp \left(x\ln \frac {x+1}{x}\right)=$$ $$=\exp (\,1-[x^{-1}/2] [1+F(x)]\,)=$$ $$=e\cdot \exp [-x^{-1}/2]\cdot\exp [-x^{-1}F(x/2)].$$ Apply the power series for $\exp$ to the 2nd & 3rd terms in the line above, to get $$\exp [-x^{-1}/2]\cdot \exp[-x^{-1}F(x)/2]=1-x^{-1}/2+x^{-1}G(x)$$ where $G(x)\to 0$ as $x\to \infty.$

So we obtain the well-known result that $$\left(\frac {x+1}{x}\right)^x=e(\,1-x^{-1}/2+x^{-1}G(x)\,)$$ where $G(x)\to 0$ as $x\to \infty.$ Taking reciprocals of this, we have $$\left(\frac {x}{x+1}\right)^x=e^{-1}(\,1+x^{-1}/2+x^{-1}H(x)\,)$$ where $H(x)\to 0$ as $x\to \infty.$

And the rest is easy.