L'Hôpital's Rule
Assuming that the following conditions are true:
- $f(x)$ and $g(x)$ must be differentiable
- $\frac{d}{dx}g(x)\neq 0$
- $\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{0}{0}\mbox{ or }\lim\limits_{x\to c} \frac{f(x)}{g(x)}= \frac{\pm\infty}{\pm\infty}$
Then,
$$ \lim\limits_{x\to c} \frac{f(x)}{g(x)}= \lim\limits_{x\to c} \frac{\frac{d}{dx}f(x)}{\frac{d}{dx}g(x)}=L $$
Where $c$ and $L$ is any real number or $\pm\infty$.
So to answer your questions, yes, L'Hôpital's rule can be used repeatedly, provided that all of the above conditions are met. Since your example doesn't meet the aforementioned conditions, L'Hôpital's rule is not applicable.
Here is a case where L'Hôpital's rule is applicable multiple times,
$$
\lim_{x\to \infty}\dfrac{e^x}{x^2}=\frac{\infty}{\infty}
$$
Since the conditions are met, we can apply L'Hôpital's rule
$$
\lim_{x\to \infty} \frac{\frac{d}{dx}e^x}{\frac{d}{dx}x^2}= \lim_{x\to \infty} \frac{e^x}{2x}= \frac{\infty}{\infty}
$$
Notice that the conditions are met again, so now
$$
\lim_{x\to \infty} \frac{\frac{d}{dx}e^x}{\frac{d}{dx}2x}= \lim_{x\to \infty} \frac{e^x}{2}= \infty
$$
Therefore
$$
\lim_{x\to \infty}\frac{e^x}{x^2}=\infty
$$
I am going to calculate your limit without using Taylor series but only the two following notable limits:
$\lim\limits_{x\to0}\dfrac{\ln(1+x)}{x}=1\;,\quad\lim\limits_{x\to0}\dfrac{x-\ln(1+x)}{x^2}=\dfrac12\;.$
$\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]=$
$=\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^2\ln\left(1+\dfrac1x\right)^x\right]=$
$=\lim\limits_{x\to\infty}\dfrac{\left(1+\frac1x\right)^x-e\ln\left(1+\frac1x\right)^x}{\left[1-\ln\left(1+\frac1x\right)^x\right]^2}\cdot\lim\limits_{x\to\infty}\left[x-x\ln\left(1+\frac1x\right)^x\right]^2$
Now I am going to calculate the first limit by using the following substitution:
$t=\dfrac1e\left(1+\dfrac1x\right)^x-1\;.$
$\lim\limits_{x\to\infty}\dfrac{\left(1+\dfrac1x\right)^x-e\ln\left(1+\dfrac1x\right)^x}{\left[1-\ln\left(1+\dfrac1x\right)^x\right]^2}=$
$=\lim\limits_{t\to0}\dfrac{e(1+t)-e\ln\big[e(1+t)\big] }{\left\{1-\ln\big[e(1+t)\big]\right\}^2}=$
$=e\cdot\lim\limits_{t\to0}\dfrac{1+t-1-\ln(1+t)}{\big[1-1-\ln(1+t)\big]^2}=$
$=e\cdot\lim\limits_{t\to0}\dfrac{t-\ln(1+t)}{\ln^2(1+t)}=$
$=e\cdot\lim\limits_{t\to0}\dfrac{t-\ln(1+t)}{t^2}\cdot\lim\limits_{t\to0}\left[\dfrac{t}{\ln(1+t)}\right]^2=$
$=e\cdot\dfrac12\cdot1^2=\dfrac e2\;.$
Now I am going to calculate the second limit by using the following substitution:
$u=\dfrac1x\;.$
$\lim\limits_{x\to\infty}\left[x-x\ln\left(1+\dfrac1x\right)^x\right]^2=$
$=\lim\limits_{x\to\infty}\left[x-x^2\ln\left(1+\dfrac1x\right)\right]^2=$
$=\lim\limits_{u\to0}\left[\dfrac1u-\dfrac1{u^2}\ln\big(1+u\big)\right]^2=$
$=\lim\limits_{u\to0}\left[\dfrac{u-\ln(1+u)}{u^2}\right]^2=$
$=\left(\dfrac12\right)^2=\dfrac14\;.$
Consequently,
$\lim\limits_{x\to\infty}\left[x^2\left(1+\dfrac1x\right)^x-ex^3\ln\left(1+\dfrac1x\right)\right]=$
$=\dfrac e2\cdot\dfrac14=\dfrac e8\;.$
Best Answer
You were going the right way. All that is left is to write the limit with $\frac 1x$ in the denominator, then one application of L'Hospital's rule.