Calculate $\lim _{x\to \infty }\left(x^3\left(1+\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)^{\frac{1}{x}}-x^3-ex\right)$ with Taylor expansion $L=\lim _{x\to \infty }\left(x^3\left(1+\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)^{\frac{1}{x}}-x^3-ex\right)$
If I do $u=\frac{1}{x} \Rightarrow x=\frac{1}{u}$
Then $L=\lim _{u\to 0 }\left(\frac{1}{u^3}\left(1+u\left(1+u\right)^\frac{1}{u}\right)^{u}-\frac{1}{u^3}-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(\left(1+u\left(1+u\right)^\frac{1}{u}\right)^{u}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+ue^{\frac{1}{u}ln(1+u)})}-1\right)-\frac{e}{u}\right)$
I know that, when $t\to0, \ln(1+t)=1-\frac{t^2}{2}+\frac{t^3}{3}-\frac{t^4}{4}+o(t^4)$
And also, $e^t=1+t+\frac{t^2}{2}+\frac{t^3}{6}+o(t^3)$
So: $L=\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+ue^{\frac{1}{u}(u-\frac{u^2}{2}-\frac{u^3}{3}-\frac{u^4}{4})})}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+ue^{(1-\frac{u}{2}-\frac{u^2}{3}-\frac{u^3}{4})})}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+u*e*e^{\frac{-u}{2}}*e^{\frac{u^2}{3}}*e^{\frac{-u^3}{4}})}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+u*e*(1-\frac{u}{2}+\frac{u^2}{8}-\frac{u^3}{48})*(1+\frac{u^2}{3})*(1-\frac{u^3}{4})))}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+e*(u-\frac{u^2}{2}+\frac{u^3}{8})*(1+\frac{u^2}{3})*(1-\frac{u^3}{4})))}-1\right)-\frac{e}{u}\right)=
\lim _{u\to 0 }\left(\frac{1}{u^3}\left(e^{u\ln(1+e*(u-\frac{u^2}{2}+\frac{11u^3}{24})}-1\right)-\frac{e}{u}\right)
$
Note: I am not writing the term $o(u^3)$ I will do it at the end
How would I do it easily using the Taylor expansion?
 A: Composing Taylor series is a piece of cake if you are patient, going from inside to outside.
$$A=x^3\left(1+\frac{1}{x}\left(1+\frac{1}{x}\right)^x\right)^{\frac{1}{x}}-x^3-ex$$ As you properly did, let $x=\frac 1u$ to get
$$A=\frac{\left(1+u (u+1)^{\frac{1}{u}}\right)^u-e u^2-1}{u^3}$$ Now, let us work the pieces
$$a=(u+1)^{\frac{1}{u}}\implies \log(a)={\frac{1}{u}}\log(1+u)=1-\frac{u}{2}+\frac{u^2}{3}-\frac{u^3}{4}+O\left(u^4\right)$$
$$a=e^{\log(a)}=e-\frac{e u}{2}+\frac{11 e u^2}{24}-\frac{7 e u^3}{16}+O\left(u^4\right)$$
$$b=1+u a=1+e u-\frac{e u^2}{2}+\frac{11 e u^3}{24}-\frac{7 e u^4}{16}+O\left(u^5\right)$$
$$c=b^u \implies \log(c)=u \log(b)=e u^2-\frac{e (1+e)}{2}  u^3+\frac{e(11 +12 e+8 e^2}{24}  u^4+O\left(u^5\right)$$
$$c=e^{\log(c)}=1+e u^2-\frac{e (1+e)}{2}  u^3+\frac{e(11+24 e+8 e^2)}{24} 
   u^4+O\left(u^5\right)$$
Finally
$$A=-\frac{e (1+e)}{2} +\frac{e(11+24 e+8 e^2)}{24} u+O\left(u^2\right)$$ which gives the limit and also how it is approached.
Try its for $u=0.01$ (quite  far away from $0$). The exact result is $-4.90454$ while the above truncated expansion gives $-4.90037$.
This means that, if you have to solve for $u$ the equation $A=-5$, the above would give as an estimate
$$u_0=\frac{12 \left(-10+e+e^2\right)}{e \left(11+24 e+8 e^2\right)}\approx 0.003501$$ while the exact solution is $0.003536$
