How to show $\lim_{x\to\infty} \frac{5^{x} - 4^{x}}{3^{x} - 2^{x}} = \infty$

Question

$$\lim_{x\to\infty} \frac{5^{x} - 4^{x}}{3^{x} - 2^{x}} = \infty$$

I can tell this is $\infty$ eventually the numerator becomes really big, and the denominator becomes really small, and that equals $\infty$.

However, I'm just guessing. How would I solve this limit to show it is indeed $\infty$. What I tried doing is rewriting the limit as:

$$\lim_{x\to\infty} \frac{e^{x \ln(5)} - e^{x \ln(4)}}{e^{x \ln(3)} - e^{x \ln(2)}} = \frac{\infty - \infty}{\infty - \infty}$$

which is not really helpful. How do I show that this limit goes to $\infty$?

Answer 1

Hint:$$\frac{5^x-4^x}{3^x-2^x}=\frac{5^x}{3^x}\cdot\frac{1-(4/5)^x}{1-(2/3)^x}$$

Answer 2

$\forall k\exists x$ such that $3^x-2^x>k$ so it's not that the denominator gets small, it's that the numerator grows faster than the denominator. We can make this explicit by factoring out the dominating terms to get

$$\left(\frac{5}{3}\right)^x\frac{1-(4/5)^x}{1-(2/3)^x}$$

The first factor diverges and the second tends to $1$.

Answer 3

Note that $5^4=625\gt2\times 4^4=256$ so that it is easy to show $5^x\gt 5^4\cdot 4^{x-4}\gt2\times 4^x$ for $x\ge 4$. We also have $3^x-2^x\lt 3^x$ for $x\gt 0$.

So for $x\ge 4$ we have $$\frac {5^x-4^x}{3^x-2^x}\gt \frac{2\cdot 4^x-4^x}{3^x}=\left(\frac 43\right)^x$$With $\frac 43\gt 1$ this is then easy.