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I see the proccedure of:

How I find the limit of $\frac{2^n}{e^{p(n)l}}$

I didn’t understand how it applies in my case:

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

$$ \lim_{x\to\infty} \frac{\ln(3)^{x}}{\ln(e^{x})}$$

$$ \lim_{x\to\infty} \ln(3)^{x} - \lim_{x\to\infty}\ln(e^{x})$$

$$ \lim_{x\to\infty} x\ln 3 - \lim_{x\to\infty}x\ln e$$

$$ \lim_{x\to\infty} x\ln 3 - \lim_{x\to\infty}x\cdot 1$$

$$ \infty\cdot\ln 3 - \infty\cdot 1$$

$$ \infty\cdot\ln 3 - \infty\cdot 1$$

$$ \infty\cdot1.098 - \infty\cdot 1\approx0$$

I don’t know if it is wrong but the result is not $$ 1 $$

I suppose applying Neperian logarithm on everything is arbitrary.

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    $\begingroup$ Note that $\frac{2}{e}<1$, but $\frac{3}{e}>1$. So it cannot apply to your case. $\endgroup$ Commented Oct 24 at 17:04
  • $\begingroup$ How can $\infty\cdot1.098-\infty\cdot1$ approximately be $0$? You just need to notice that $\dfrac{3}{\mathrm{e}}>1$, and $\lim\limits_{n\to\infty}a^n=+\infty$ for $a>1$. $\endgroup$ Commented Oct 24 at 17:04
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    $\begingroup$ One thing to keep in mind is what you learn from the study of indeterminate forms: the moment you reach an expression of the form $\infty - \infty$ you know that you have an indeterminate form, and so you must stop and do something else. $\endgroup$ Commented Oct 24 at 17:05
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    $\begingroup$ You got $\infty-\infty$, which doesn't have an associated value. Instead, following the spirit of your initial idea, you could do $\lim_{x\to+\infty}\ln(3^x/3^x)=\lim_{x\to+\infty}x\ln(3/e)=+\infty\cdot \ln(3/e)$. Since $\ln(3/e)>0$, then $+\infty\cdot \ln(3/e)=+\infty$. $\endgroup$ Commented Oct 24 at 17:08
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    $\begingroup$ For any $~m \in \mathbb{R_{> 1}},~$ such as $~m = 3/e,~$ you have that $$\lim_{x \to +\infty} ~m^x = +\infty.$$ This can be routinely proven by (among other methods), using logarithms. Once you have proven this, the analysis in the answer of Afntu should fall into place. $\endgroup$ Commented Oct 25 at 7:49

1 Answer 1

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Let $a = 3/e>1$, then Since $a>1$, there is $\varepsilon>0$ s.t. $a=1+\varepsilon.$ Then $a^n=(1+\varepsilon)^n\geq n\varepsilon+1$, so $\displaystyle\lim_{ n\to \infty} a^n = \infty$.

Otherwise, $a^x$ is an increasing continuous function for $a>1$, hence also the result.

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    $\begingroup$ Moderate alternative approach: $$f(x) = a^x = e^{x \ln(a)} ~: a > 1 \implies $$ $$f'(x) = \ln(a) \times a^x ~: ~\ln(a) > 0.$$ Since $~\ln(a)~$ is a fixed positive constant, and $~a^x~$ is a strictly increasing function, this implies that $~f'(x)~$ itself is increasing at a larger and larger rate, as $~x \to +\infty.~$ ~Therefore, $~f(x),~$ which is a continuous function, must be unbounded. $\endgroup$ Commented Oct 25 at 7:58

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