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.

Multiplying by $\displaystyle\frac{x e^{\frac{1}{x}} + \sqrt{x^2+x+1}}{x e^{\frac{1}{x}} + \sqrt{x^2+x+1}}$ we have 
$$\lim_{x\to\infty}\left(x e^{\frac{1}{x}} - \sqrt{x^2+x+1} \right)=\lim_{x\to\infty}\frac{x^2 (e^{\frac{2}{x}} -1)-x-1 }{x e^{\frac{1}{x}} + \sqrt{x^2+x+1} }=\lim_{x\to\infty}\frac{x (e^{\frac{2}{x}} -1)-1-1/x }{ e^{\frac{1}{x}} + \sqrt{1+1/x+1/x^2} }.$$
Now $\displaystyle\lim_{x\to\infty}x(e^{\frac{2}{x}} -1)=\lim_{x\to\infty}\frac{e^{\frac{2}{x}} -1}{1/x}=\lim_{x\to\infty}\frac{e^{\frac{2}{x}}(-2/x^2)}{-1/x^2}=2$. 
Thus the final limit is: $\frac{2-1}{1+\sqrt{1}}=1/2$.
