<s>[INST] Calculating $\lim_{x\to\infty}\left(x e^{\frac{1}{x}} - \sqrt{x^2+x+1} \right)$ I've managed to solve it by rewriting the expression as
$$\frac{1 - \frac{\sqrt{x^2 +x + 1}}{x e^{\frac{1}{x}}} }{ \frac{1}{x e^{\frac{1}{x}}} }$$
then applying L'Hospital's rule.
This took up one whole page and was very hairy, even after substituting $t = \sqrt{x^2+x+1}$. I'm wondering if there's a simpler way. A friend suggested substituting $e = (1+\frac{1}{x})^x$, but that's a bit suspicious.
In both cases, the answer is $\frac{1}{2}$, as confirmed by my computer.
 [/INST] Using the standard limit
$$
\lim_{t\to 0}\frac{e^t-1}{t}=1
$$
we have 
$$
1=\lim_{x\to\infty}\frac{e^{1/x}-1}{1/x}=\lim_{x\to\infty}(xe^{1/x}-x).
$$
Now rewrite your limit as
$$
\lim_{x\to\infty}(xe^{1/x}-x+x-\sqrt{x^2+x+1})=1+\lim_{x\to\infty}(x-\sqrt{x^2+x+1}).
$$
To calculate the last limit we rewrite again
\begin{align}
&\lim_{x\to\infty}(x-\sqrt{x^2+x+1})=\lim_{x\to\infty}\frac{(x-\sqrt{x^2+x+1})(x+\sqrt{x^2+x+1})}{x+\sqrt{x^2+x+1}}=\\
&=\lim_{x\to\infty}\frac{x^2-(x^2+x+1)}{x+\sqrt{x^2+x+1}}=
\lim_{x\to\infty}\frac{-x-1}{x+\sqrt{x^2+x+1}}=\lim_{x\to\infty}\frac{-1-\frac1x}{1+\sqrt{1+\frac1x+\frac{1}{x^2}}}=-\frac{1}{2}.
\end{align}
</s>