Why does $\lim_{x \to 0}(x^2+x+1)^{1/x}$ equal $e$? 
Why does  $$\lim_{x \to 0}\Big(x^2+x+1\Big)^{\large 1/x} =e \;\;?$$

I only got a brief explanation that basically the equation brings itself to equal e, but I didn't catch the steps involved. I know it involves L'Hopital's rule, but if someone could do a step by step I'd appreciate it.
 A: Hint:
Let
\begin{align*}
y & = (1+x+x^2)^{1/x}\\
\ln y & = \frac{\ln(1+x+x^2)}{x}\\
\lim_{x \to 0}\ln y & = \lim_{x \to 0}\frac{\ln(1+x+x^2)}{x}\\
& = \lim_{x \to 0}\frac{\frac{1}{1+x+x^2}(1+2x)}{1}
\end{align*}
Hope you can take it from here.
A: Hints: Notice that 
$$ \lim_{x \to 0^+} \frac{ \ln (x^2 + x + 1) }{x } = \lim_{x \to 0^+} \frac{\frac{2x+1}{x^2+x+1}}{1} = 1 $$
And notice that 
$$ (x^2+x+1)^{1/x} = \exp \ln ( x^2+x+1)^{1/x} = \exp \left( \frac{ \ln (x^2+x+1) }{x} \right)$$
A: I assume as known that


*

*$\lim_{t\to 0}\left( 1+t\right)^{\frac{1}{t}} = e$ 

*Consequently, you have $\lim_{x\to 0}\left(1 + \color{blue}{x+x^2} \right)^{\frac{1}{\color{blue}{x+x^2}}} = e$
Now, you can write
$$\begin{eqnarray*} \Big(x^2+x+1\Big)^{\large 1/x}
 & = & \left(1 + \color{blue}{x+x^2} \right)^{\frac{1}{\color{blue}{x+x^2}}} \cdot \left(1 + \color{blue}{x+x^2} \right)^{\frac{\color{blue}{x+x^2}}{x}} \\
 & = & \underbrace{\left(1 + \color{blue}{x+x^2} \right)^{\frac{1}{\color{blue}{x+x^2}}}}_{\stackrel{x\to 0}{\longrightarrow}e} \cdot \underbrace{\left(1 + \color{blue}{x+x^2} \right)^{1+x}}_{\stackrel{x\to 0}{\longrightarrow}1^1} \\
& \stackrel{x\to 0}{\longrightarrow} & e
\end{eqnarray*}$$
A: We don't need l'Hopital indeed by $x\to 0^+,\, y=\frac1x \to \infty$
$$\lim_{x \to 0}\Big(x^2+x+1\Big)^{\large 1/x}=\lim_{y \to \infty}\Big(1+\frac1y+\frac1{y^2}\Big)^{y} =\lim_{y \to \infty}\left[\Big(1+\frac1y+\frac1{y^2}\Big)^{\frac1{\frac1y+\frac1{y^2}}}\right]^{1+\frac1{y}}\to e$$
and silimilarly for the case $x\to 0^-$.
