Justifying the result of $\lim_{x\to\infty} \tfrac{3^x}{4^x}$ I am working with the next limit:
$$\lim_{x\to\infty} \frac{3^x}{4^x}$$
I intuitively know that since
$$3^x< 4^x$$
when $x$ tends to infinite,
the result of the limit is:
$$\lim_{x\to\infty} \frac{3^x}{4^x}=0$$
However, I need a some more mathematical justification rather than the intuitive justification, I would appreciate any help or hint to justify this result, the l'Hospital's rule doesn't work for this limit, since if I apply the rule, the limit remains similar.
 A: Hint: $$\frac{3^x}{4^x}=\left(\frac34\right)^x$$and use the fact that $\frac34<1$.
A: $$\lim_{x\to\infty} \frac{3^x}{4^x}=\lim_{x\to\infty} \left(\frac{3}{4}\right)^x=0$$
Because $\left(\tfrac34\right)^x$ is decreasing to $0$, as exponential function $a^x$ with $a=\tfrac34<1$.
A: Just work out the division: it's $(3/4)^x$ which is a decreasing exponential.
A: Say
$$y=\lim_{x\to\infty}\frac{3^x}{4^x}$$
We then have
$$y=\lim_{x\to\infty}\left(\frac{3}{4}\right)^x$$
We then take the natural log of both sides:
$$\ln y=\ln\left(\lim_{x\to\infty}\left(\frac{3}{4}\right)^x\right)=\lim_{x\to\infty}\ln\left(\frac{3}{4}\right)^x=\lim_{x\to\infty}x\ln\left(\frac{3}{4}\right)$$
Since
$$\frac{3}{4}<1\text{ and }\ln x <0 \text{ if }x<1$$
As $x\to\infty$, the right side of the equation goes to $-\infty$. Taking
$$y=e^{\ln y}=\lim_{x\to\infty}e^{-x}$$
gets us
$$y=0$$
This isn't intuitive, but it's certainly mathematically justifiable.
A: $$\lim_{x\to\infty}\space\frac{3^x}{4^x}=\lim_{x\to\infty}\space\left(\frac{3}{4}\right)^x=\lim_{x\to\infty}\space\exp\left(\ln\left(\left(\frac{3}{4}\right)^x\right)\right)=$$
$$\lim_{x\to\infty}\space\exp\left(x\ln\left(\frac{3}{4}\right)\right)=\exp\left(\lim_{x\to\infty}\space x\ln\left(\frac{3}{4}\right)\right)$$
$$\exp\left(\ln\left(\frac{3}{4}\right)\lim_{x\to\infty}\space x\right)=\exp\left(-\ln\left(\frac{4}{3}\right)\lim_{x\to\infty}\space x\right)=0$$

With some strange looking math ($\infty$ isn't a number)
$$\exp\left(-\infty\right)=e^{-\infty}=\frac{1}{e^{\infty}}=0$$
