Finding $\lim_{n \to \infty} \frac{2^{n+1} + 3^{n+1}}{2^{n}+3^{n}}$

Question

I need help with finding

$\displaystyle \lim_{n \to \infty} \frac{2^{n+1} + 3^{n+1}}{2^{n}+3^{n}}$

Thanks!

Answer 1

Since $$2^n=_\infty o(3^n)$$

$$ \lim_{n \to \infty} \frac{2^{n+1} + 3^{n+1}}{2^{n}+3^{n}} =\lim_{n \to \infty} \frac{ 3^{n+1}}{3^{n}}=3$$

Answer 2

Consider

$$\lim_{n \to \infty} \frac{2^{n+1} + 3^{n+1}}{2^{n}+3^{n}} =\lim_{n \to \infty} \frac{2 \left(\frac{2}{3}\right)^{n} + 3}{\left(\frac{2}{3}\right)^{n}+1} $$

Now apply

$$\lim_{n\to \infty}x^n =0 \,\,\, |x|<1$$

Answer 3

Here is how you proceed $$ \displaystyle \lim_{n \to \infty} \frac{2^{n+1} + 3^{n+1}}{2^{n}+3^{n}}= \displaystyle \lim_{n \to \infty} \frac{3^{n+1}((2/3)^{n+1} + 1)}{3^n((2/3)^{n}+1)} = 3.$$

Note:

Since $\lim_{n\to \infty} \frac{2^n}{3^n}=0$ the $2^n=o(3^n)$.

Answer 4

$$ \lim_{n \to \infty} \frac{2^{n+1} + 3^{n+1}}{2^{n}+3^{n}} = \lim_{n \to \infty} \frac{(\frac{2}{3})^{n+1} + 1}{(\frac{1}{3})(\frac{2}{3})^{n}+\frac{1}{3}} = \frac{\,\,1\,\,}{\frac{1}{3}} = 3 $$