<s>[INST] Find $\lim\limits_{x \to\infty} (\frac {2+3x}{2x+1})^{x+1}$ without L'Hopital $$
\lim_{x\to\infty}\left(\frac {2+3x}{2x+1}\right)^{x+1}
$$
Not sure how to deal with this, I've tried doing the following
$$
\lim_{x\to\infty}\left(\frac {2+3x}{2x+1}\right)^{x}\cdot \lim_{x\to\infty}\left(\frac{2+3x}{2x+1}\right)
$$
Then I tried dividing by $x,$ but my teacher told me it was wrong.
 [/INST] We have 
\begin{align}
\lim_{x\to\infty}\left(\frac {2+3x}{1+2x}\right)^{1+x}
&=
\lim_{x\to\infty}\left(\frac {(1+2x)+(1+x)}{1+2x}\right)^{1+x}
\\
&=
\lim_{x\to\infty}\left(1+\frac {1+x}{1+2x}\right)^{1+x}
\\
&=
\lim_{x\to\infty}\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{1+x}
\\
&=
\lim_{x\to\infty}\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x}
\\
\end{align}
For $ x $ large enough we have
$$
1.3<\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)<3
$$
and 
$$
1.3^{1+2x}
<\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x}
<3^{1+2x}
$$
By the sandwich theorem we have to
$$
\lim_{x\to \infty}\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x}=\infty
$$
</s>