Evaluate

$\begin{aligned} L & = \lim_{n \to \infty} \frac {2^{n+1}+3^{n+1}}{2^n+3^n} & \small \color{#3D99F6} \text{Dividing up and down by }3^n \\ & = \lim_{n \to \infty} \frac {\frac {2^{n+1}}{3^n}+\frac {3^{n+1}}{3^n}}{\frac {2^n}{3^n}+\frac {3^n}{3^n}} \\ & = \frac {0+3}{0+1} = \boxed{3} \end{aligned}$

If we examine the graph of a^x the slope of 3^x is greater than 2^x and 3^x>>2^x,therefore 2^x can be neglected and left will be 3^{n+1}/3^n. therefore answer is 3. another way,we can apply binomial expansion in (3-1)^{n+1} and (3-1)^n.and then it can be evaluated.