Evaluate:- $$\lim_{x\to\infty}\left[ x\left(1+\frac1x\right)^x-kx^2\ln\left(1+\frac1x\right)\right]$$
I tried calculating the limits of the two terms separately. By applying L'Hopital, the 2nd limit can be solved by writing $x^2\ln\left(1+\frac1x\right)=\frac{\ln\left(1+\frac1x\right)}{\frac{1}{x^2}}$. Now, L'Hopital gives the final form as $\frac{x^2}{2(1+x)}=\infty$. Now, for the 1st term, we know $\left(1+\frac1x\right)^x$ is $e$ as $x$ goes to infinity, so writing the 1st term as $\frac{\left(1+\frac1x\right)^x}{\frac{1}{x}}$, we get $\frac{e}{0}=\infty$. Now, since both limits are infinite, how can we determine the actual limit?
As $x\to+\infty$, $u:=x\ln\left(1+\frac1x\right)-1\sim-\frac1{2x}\to0$, and
$$\begin{align}\left(1+\frac1x\right)^x-kx\ln\left(1+\frac1x\right)&=e^{1+u}-k(1+u)\\&=(e-k)(1+u)+\frac e2u^2+o(u^2)\\ &\begin{cases}\to e-k&\text{if }e\ne k\\\sim\frac e{8x^2}&\text{if }e=k \end{cases} \end{align}$$ hence $$\lim_{x\to\infty}x\left(\left(1+\frac1x\right)^x-kx\ln\left(1+\frac1x\right)\right)=\begin{cases}+\infty&\text{if }e>k\\-\infty&\text{if }e<k\\0&\text{if }e=k. \end{cases} $$