Limit as $x$ approaches infinity of a ratio of exponential functions.

Question

The problem appears as follows:

Use algebra to solve the following:

$$\lim_{x\to \infty} \frac{2^{3x+2}}{3^{x+3}}$$

The result is infinity which makes intuitive sense but I can’t get it to yield algebraically.

Attempt:

$$\frac{2^{3x+2}}{3^{x+3}} = \frac{2^{3x}2^2}{3^{x}3^3}$$

I’ve decided against continuing writing out my attempt because writing on an iPad, in Latex, is no fun.

The end-result I produce when taking the limit (as $x$ approaches infinity) of the above expression is $\inf / \inf = 1$.

(Do I have to demonstrate that $2^3x$ grows faster than $3^x$ for large $x$?)

Answer 1

Hint: $\displaystyle\;\frac{2^{3x+2}}{3^{x+3}}=\frac{2^2}{3^3} \cdot \frac{\left(2^3\right)^x}{3^x}=\frac{4}{27} \cdot \left(\frac{8}{3}\right)^x \gt \frac{4}{27} \cdot 2^x\,$.

Answer 2

$$\frac{2^{3x}}{3^x}=\left(\frac{2^3}{3}\right)^x=\left(\frac{8}{3}\right)^x\to \infty \text{ as } x\to \infty$$

Answer 3

Since \begin{align*} \lim_{x\to \infty} \frac{2^{3x+2}}{3^{x+3}} &= \lim_{x\to \infty} \frac{2^2 (2^3)^x}{3^3 3^x} \\ &= \lim_{x\to \infty} \frac{2^2}{3^3}\left( \frac{8}{3} \right)^x \\ &= \frac{2^2}{3^3} \lim_{x\to \infty} \left( \frac{8}{3} \right)^x \\ &> \frac{2^2}{3^3} \lim_{x\to \infty} 2^x, \\ \end{align*} we know that $\displaystyle{\lim_{x\to \infty}} 2^x=\infty$. So $\displaystyle{\lim_{x\to \infty}} \frac{2^{3x+2}}{3^{x+3}} =\infty$.