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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.

Kamil Jarosz
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Neo
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  • Knowing that $f(x) < g(x)$ does not imply that $\lim\limits_{x \rightarrow \infty} \frac{f(x)}{g(x)} = 0$. For example $x < x + 1$, but $\lim\limits_{x \rightarrow \infty} \frac{x}{x + 1} = 1$. All you can say is that if $f(x) < g(x)$ (for all sufficiently large values of $x$) and the limit $\lim\limits_{x \rightarrow \infty} \frac{f(x)}{g(x)}$ exists, then $\lim\limits_{x \rightarrow \infty} \frac{f(x)}{g(x)} \leq 1$. – Michael Joyce Nov 29 '15 at 21:49

5 Answers5

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Hint: $$\frac{3^x}{4^x}=\left(\frac34\right)^x$$and use the fact that $\frac34<1$.

Clayton
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$$\lim_{x\to\infty} \frac{3^x}{4^x}=\lim_{x\to\infty} \left(\frac{3}{4}\right)^x=0$$

Because $\left(\tfrac34\right)^x$ is decreasing to $0$, as exponential function $a^x$ with $a=\tfrac34<1$.

Kamil Jarosz
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1

Just work out the division: it's $(3/4)^x$ which is a decreasing exponential.

Justpassingby
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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.

HDE 226868
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$$\lim_{x\to\infty}\space\frac{3^x}{4^x}=\lim_{x\to\infty}\space\left(\frac{3}{4}\right)^x=\lim_{x\to\infty}\space\exp\left(\ln\left(\left(\frac{3}{4}\right)^x\right)\right)=$$ $$\lim_{x\to\infty}\space\exp\left(x\ln\left(\frac{3}{4}\right)\right)=\exp\left(\lim_{x\to\infty}\space x\ln\left(\frac{3}{4}\right)\right)$$ $$\exp\left(\ln\left(\frac{3}{4}\right)\lim_{x\to\infty}\space x\right)=\exp\left(-\ln\left(\frac{4}{3}\right)\lim_{x\to\infty}\space x\right)=0$$

With some strange looking math ($\infty$ isn't a number)

$$\exp\left(-\infty\right)=e^{-\infty}=\frac{1}{e^{\infty}}=0$$

Jan Eerland
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