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.
 A: It is $\infty,$ because $\frac{2+3x}{2x+1} \to 3/2$ and $x+1 \to \infty.$
A: The inner fraction converges to $\frac 3 2$, which is larger than 1,  so the whole thing goes to $+\infty$.
A: Write $(\frac {2+3x}{2x+1})^{x+1}=e^{\ln{(\frac {2+3x}{2x+1})^{x+1}}}=$  $ e^{(x+1)\ln(\frac{2+3x}{2x+1}) }$ ...
A: HINT
We have that 
$$\left(\frac {2+3x}{2x+1}\right)^{x+1}\sim \left(\frac32\right)^{x+1}$$
A: When $x\to\infty$: $$(\frac{2+3x}{2x+1})^{x+1}=(\frac{3t-1}{2t-1})^t~~~\text{for}~~ (x+1=t)$$ $$=(3/2+\frac{1/2}{2t-1})^t=(3/2+\frac{1/2}{s})^{s}\times3/2$$ where  $s=2t-1$ and the last parentheses goes to $+\infty$. This is the same as @gimusi noted.
A: 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
$$
\lim_{x\to \infty}\left[\left(1+\frac {1}{\frac{1+2x}{1+x}}\right)^{\frac{1+x}{1+2x}}\right]^{1+2x}=\infty
$$
A: You get the answer by dividing the numerator and the denominator by x where x is not 0.
the numerator witll give you 2/x+3 and the lim when x+1 goes to infinity is 3. 
The same thing with the denominator will give you 2+1/x and the lim of that is 2 
The answer is 3/2.
A: $$L=\lim_{x\to \infty}\left(\frac{2+3x}{2x+1}\right)^{x+1}$$
we can see that as $x\to \infty$ the $2$ at the top and $1$ at the bottom become arbitary small, such that we can approximate:
$$\lim_{x\to \infty}\left(\frac{2+3x}{2x+1}\right)=\lim_{x\to \infty}\left(\frac{3x}{2x}\right)=\frac{3}{2}$$
so it can also be said that:
$$\lim_{x\to \infty}\left(\frac{2+3x}{2x+1}\right)^{x+1}=\lim_{x\to \infty}\left(\frac{3}{2}\right)^{x+1}$$
and since $\frac{3}{2}>1$, this tends to infinity, so the limit is not convergent.
