Solution verification: $\lim_{x\to \infty}\Bigg(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\Bigg)^{x\ln x}$ 
Find: $$\displaystyle\lim_{x\to
 \infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}$$

My attempt:
$\displaystyle\lim_{x\to\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln(x^2+3x+4)-\ln(x^2+2x+3)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\\\displaystyle\lim_{x\to\infty}\left(1+\frac{\ln\left(\frac{x^2+3x+4}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}=\lim_{x\to\infty}\left(1+\frac{\ln\left(1+\frac{x+1}{x^2+2x+3}\right)}{\ln(x^2+2x+3)}\right)^{x\ln x}$
What I used:
$\displaystyle\lim_{x\to\infty}\ln\left(1+\frac{x+1}{x^2+2x+3}\right)=0\;\;\&\;\;\lim_{x\to\infty}\ln(x^2+2x+3)=+\infty$
In the end, I got an indeterminate form: $\displaystyle\lim_{x\to\infty}1^{x\ln x}=1^{\infty}$
Have I made a mistake anywhere? It seems suspicious.
added: replacement $\frac{x+1}{x^2+2x+3}$ by $\frac{1}{x}$ wasn't appealing either.
Would:$$\lim_{x\to\infty}\Big(\Big(1+\frac{1}{x}\Big)^x\Big)^{\ln x}=x=\infty$$
be wrong?
//a few days after users had provided hints and answered the question,we discussed this with our assistant and he suggested a table formula that can also be applied (essentialy the last step in methods provided in the answers I recieved).:$$\lim_{x\to c}f(x)=1\;\&\;\lim_{x\to c}g(x)=\pm\infty$$then$$\lim_{x\to c}f(x)^{g(x)}=e^{\lim_{x\to c}(f(x)-1)g(x)}//$$

Use $\lim_{t\to 1}\frac{\ln(t)}{t-1}=1$ twice: first for $t=\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}$ and then for $t=\frac{x^2+3x+4}{x^2+2x+3}$. The rest is simplification.
$$\begin{align}
\lim_{x\to\infty}\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)^{x\ln x} &=
\exp\left(\lim_{x\to\infty}\ln\left(\frac{\ln(x^2+3x+4)}{\ln(x^2+2x+3)}\right)x\ln x\right) \\
&= \exp\left(\lim_{x\to\infty}\frac{\ln\frac{x^2+3x+4}{x^2+2x+3}}{\ln(x^2+2x+3)}x\ln x\right) \\
&=\exp\left(\lim_{x\to\infty}\frac{(x+1)x}{x^2+2x+3}\frac{\ln x}{\ln(x^2+2x+3)}\right)
\end{align} $$
We have
$$\frac{(x+1)x}{x^2+2x+3}=\frac{x^2+x}{x^2+2x+3}\to 1 $$
For the other fraction, use that $\ln(x^2+2x+3)=\ln(x^2(1+2/x+3/x^2))=2\ln x+\ln(1+2/x+3/x^2)$
$$\frac{\ln x}{\ln(x^2+2x+3)}=\frac{\ln x}{2\ln x+\ln(1+2/x+3/x^2)}=\frac{1}{2+\frac{\ln(1+2/x+3/x^2)}{\ln x}}\to \frac 12 $$
So the limit is $e^{\frac 12 \cdot 1}=\sqrt{e}$.
