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I am having some diffuculty finding the limit for this expression and would appreciate if anyone could give a hint, as to how to continue. I know the limit must be $e^{14}$ (trough an engine) and I can show it for

$(1+\frac{2}{x}+\frac{1}{x^2})^{7x}$ like

$\lim_{x\to \infty}(1+\frac{2}{x}+\frac{1}{x^2})^{7x} = ((1+\frac{1}{x})^2)^{7x} = (1+\frac{1}{x})^{x \cdot 7 \cdot 2} = e^{7 \cdot 2} = e^{14}$

However the $\frac{3}{x^2}$ in $(1+\frac{2}{x}+\frac{3}{x^2})^{7x}$ is causing problems for me I have:

$\lim_{x\to \infty}(1+\frac{2}{x}+\frac{3}{x^2})^{7x} = (1+\frac{2}{x}+\frac{1}{x^2} +\frac{2}{x^2})^{7x} = ((1+\frac{1}{x})^2+\frac{2}{x^2})^{7x} =...$

but I'm not sure how to continue (how to get rid of $\frac{2}{x^2}$) I was thinking that using the binomial theorem might somehow reduce the$\frac{2}{x^2}$ for $\lim_{x\to \infty}$, but I'm not sure how to do that for an x in the exponent, or if it is useful, or if it is even allowed in this case.

If anyone could point me in the right direction, as to how to continue I would be very grateful.

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Limits of the form $f(x)^{g(x)}$ are usually best approached by converting into the exponential form $\exp[g(x)\log f(x)]$. The present limit is no exception. $$ \lim_{x\to \infty}(1+\frac{2}{x}+\frac{3}{x^2})^{7x}=\lim_{x\to\infty}e^{7x \log(1+2/x+3/x^2)}\ . $$ Then use $\log(1+\epsilon)\sim \epsilon$ for $\epsilon\to 0$, to deduce that your limit is $$ \sim e^{7x(2/x+3/x^2)}=e^{14}\ . $$ Note that $2/x+3/x^2\to 0$ as $x\to\infty$, so you can safely set $\epsilon=2/x+3/x^2$ in the logarithm expansion above, as I suggested.

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  • $\begingroup$ You can also utilize l'Hopital's rule for the exponent (since it is of the form $\infty\cdot0$, it can easily be rearranged to fit either form of $0/0$ or $\infty/\infty$...this removes the necessity to argue that $\log(1+\varepsilon)\sim\varepsilon$). $\endgroup$ – Clayton Mar 7 '18 at 16:43
  • $\begingroup$ Thank you alot! I will try the other suggested solutions as well, but this was the easiest one to understand. $\endgroup$ – Viceh Mar 7 '18 at 17:01
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$$\lim_{x\to\infty}\left(1+\dfrac{mx+n}{x^2}\right)^{7x}=\left(\lim_{x\to\infty}\left(1+\dfrac{mx+n}{x^2}\right)^{x^2/(mx+n)}\right)^{\lim_{x\to\infty}7(mx+n)/x}$$

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It can be convenient to use Taylor series when doing problems like this:

$$\lim_{x \to \infty} \left(1 + \frac{2}{x} + \frac{3}{x^2}\right)^{7x} = \lim_{x \to \infty} e^{7x\ln(1+\frac{2}{x}+\frac{3}{x^2})} = \lim_{x\to \infty} e^{7x\left( \frac{2}{x} + \frac{3}{x^2} + \mathcal{O}\left(\left((\frac{2}{x} + \frac{3}{x^2}\right)^2\right) \right)} = \lim_{x\to \infty} e^{14 + \frac{21}{x}} = e^{14}$$

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  • $\begingroup$ Here we have used for small $t$ $$\ln(1+t) = \sum_{n=1}^{\infty} (-1)^{n-1} \frac{t^n}{n}$$ $\endgroup$ – MelpomenicMuse Mar 7 '18 at 16:46
  • $\begingroup$ Thanks for your Input, I definitely will try this, since I need practise with Taylor-Series aswell! $\endgroup$ – Viceh Mar 7 '18 at 17:06
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Screwing the rigo, as $x\to \infty$, $\frac{1}{x^2}$ can be neglected, and we are left with

$$\lim_{x\to +\infty}\left(1 + \frac{2}{x} + \mathcal{O}(x^{-2})\right)^{7x} = \lim_{x\to +\infty}\left(\left(1 + \frac{2}{x}\right)^x\right)^7 = (e^2)^7) = e^{14}$$

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  • $\begingroup$ as $x\to \infty$, $\frac{2}{x}$ goes to zero too, but I am happy to see you don't neglect that. $\endgroup$ – imranfat Mar 7 '18 at 16:59
  • $\begingroup$ @imranfat Ah no, it would have been a crime! $\endgroup$ – Von Neumann Mar 7 '18 at 17:00
  • $\begingroup$ I have to say I don't understand why $\frac{1}{x^2}$ can be neglected and $\frac{2}{x}$ can't be ? $\endgroup$ – Viceh Mar 7 '18 at 17:04
  • $\begingroup$ @Viceh Not really neglected. It's an asymptotic analysis, hence I treated order greater or equal than $x^{-2}$ as asymptotically zero as $x \to \infty$ $\endgroup$ – Von Neumann Mar 7 '18 at 17:09
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Note that

$$\left(1+\frac{2}{x}+\frac{3}{x^2}\right)^{7x}=\left[\left(1+ \frac{2}{x}+\frac{3}{x^2}\right)^{\frac1{\left(\frac{2}{x}+\frac{3}{x^2}\right)}}\right]^{{7x}{\left(\frac{2}{x}+\frac{3}{x^2}\right)}}\to e^{14}$$

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