Can someone help me solve this limits question? $$\begin{align}\lim x → ∞\end{align}$$

$$\begin{align}
f(x) = {\frac{2^{x+1}+{3^{x+1}}}{2^x + 3^x}} \\
\end{align}$$

I tried using L Hopitable but that gives the same expression. Also tried using substitution but I didn't get anywhere. Help would be appreciated.
Here is another idea. Try writing,
\begin{align*}
\frac{2^{x+1}+3^{x+1}}{2^x+3^x} &amp;= \frac{2 \cdot 2^x+3 \cdot 3^x}{2^x+3^x} \\
&amp;=\frac{3\cdot 2^x - 2^x+3 \cdot 3^x}{2^x+3^x} \\
&amp;=\frac{3\left( 2^x+3^x \right)-2^x}{2^x+3^x} \\
&amp;=3-\frac{2^x}{2^x+3^x} \\
&amp;=3-\frac{\frac{2^x}{2^x}}{\frac{2^x}{2^x}+\frac{3^x}{2^x}} \\
&amp;=3-\frac{1}{1+\left( \frac{3}{2} \right)^x}.
\end{align*}
Now most people would just look at this and say
$$\lim_{x \rightarrow \infty} \left( 3-\frac{1}{1+\left( \frac{3}{2} \right)^x} \right)=3.$$
To begin with
$$f(x)=\frac{2\cdot(\frac23)^x+3}{(\frac23)^x+1}$$
it follows that
$$\lim_{x\to+\infty}f(x)=3.$$
However
$$f(x)=\frac{2+3\cdot(\frac32)^x}{1+(\frac32)^x}$$
in such a case, we get that
$$\lim_{x\to-\infty}f(x)=2.$$