A limit with an exponent

I have this limit as a homework assignment, $$ \lim_{x\to+\infty} \left(\frac {x^2 + x + 1} {x +3}\right)^{\frac {1} {x^2+2x+4}} $$ I think it can be solved using the number e, but I don't know how to make it in the form where I use that. It would be of great help if you can explain to me at least the first and the second step. Thank you in advance!

First note that for big $x$ we have that $$ 1\leq\left(\frac {x^2 + x + 1} {x +3}\right)^{\frac {1} {x^2+2x+4}} \leq\left({x^2+2x+4}\right)^{\frac {1} {x^2+2x+4}}\to 1$$ for $x\to\infty$ since $x^{1/x}\to1$ for $x\to\infty$.

A different way is (using the Hint from TheSilverDoe),

\begin{align} 1\leq\left(\frac {x^2 + x + 1} {x +3}\right)^{\frac {1} {x^2+2x+4}} &= \exp\left[\frac {1} {x^2+2x+4}\ln{\left(\frac {x^2 + x + 1} {x +3}\right)}\right]\\ &\leq\exp\left[\frac {1} {x^2+2x+4}\cdot{\frac {x^2 + x + 1} {x +3}}\right]\\ &\leq\exp\left[\frac {1} {x^2+2x+4}\cdot{\frac {x^2 + 2x + 4} {x +3}}\right]\\ &\rightarrow 1 \end{align}