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I have a little problem with limit of this function:

$\lim_{x \to \infty} x^2(2017^{\frac{1}{x}} - 2017^{\frac{1}{x+1}})$

I have tried de L'Hopital rule twice, but it doesn't work. Now I have no idea how to do it.

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    $\begingroup$ Note that $\frac1x-\frac1{x+1}=\frac1{x(x+1)}$ hence this is $$\frac{u\left(\frac1x\right)-u\left(\frac1{x+1}\right)}{\frac1x-\frac1{x+1}}\cdot\frac{x}{x+1}$$ where $$u(t)=2017^t$$ By the MVT, the first ratio is $$u'(\xi_x)$$ for some $$\frac1{x+1}\leqslant\xi_x\leqslant\frac1x$$ hence, if $u'$ is continuous at $0$, the desired limit is $$u'(0)\cdot1=u'(0)$$ Can you check this continuity and compute this value? // This approach allows to to show that, for every differentiable $v$, $$x^2\left(v\left(\frac1x\right)-v\left(\frac1{x+1}\right)\right)\to v'(0)$$ provided $v'$ is continuous at $0$. $\endgroup$ – Did Jan 17 '17 at 11:55
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    $\begingroup$ @Did this is definitely a fully fledged answer. Not sure why you are keeping that in the comment section... In fact, I would argue that it's the best answer as of now. $\endgroup$ – Brevan Ellefsen Jan 17 '17 at 14:45
  • $\begingroup$ @BrevanEllefsen Thanks for the appreciation. (But the OP had already accepted an answer two hours earlier, so...) $\endgroup$ – Did Jan 17 '17 at 14:51
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    $\begingroup$ @Did fair enough. I've always been of the opinion that answers added after a good answer has been accepted must use a technique not already present in one of the answers, and must either be A) interesting enough to warrant a new post or B) gives insights not already provided by existing answers. In my mind your question fulfills both, hence my comment. $\endgroup$ – Brevan Ellefsen Jan 17 '17 at 14:58
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For finite $a>0,$

$$\lim_{x\to\infty}x^2(a^{1/x}-a^{1/(x+1)})=\lim_{x\to\infty}a^{1/(x+1)}\lim_{x\to\infty}x^2\left(a^{\{1/x-1/(x+1)\}}-1\}\right)$$

Now $\displaystyle x^2\left(a^{\{1/x-1/(x+1)\}}-1\}\right)=\dfrac{a^{\frac1{x(x+1)}}-1}{\frac1{x(x+1)}}\cdot\dfrac1{1+\frac1x}$

Set $\frac1{x(x+1)}=y$ to use $\lim_{y\to0}\dfrac{a^y-1}y=\ln a$

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Let us consider $$A=x^2(a^{\frac{1}{x}} -a^{\frac{1}{x+1}})$$ $$a^{\frac{1}{x}}=e^{\frac{\log(a)}x}=1+\frac{\log (a)}{x}+\frac{\log ^2(a)}{2 x^2}+\frac{\log ^3(a)}{6 x^3}+O\left(\frac{1}{x^4}\right)$$ Do the same for the other term; subtract from eash other, use common denominator and so on.

For the first term, $$a^{\frac{1}{x}} -a^{\frac{1}{x+1}}=\log(a)\left(\frac{1}{x}-\frac{1}{x+1} \right)+\cdots=\log(a)\frac{x+1-x}{x(x+1)}+\cdots=\frac{\log(a)}{x(x+1)}+\cdots$$ and so on.

You should arrive to $$a^{\frac{1}{x}} -a^{\frac{1}{x+1}}=\frac{\log (a)}{x^2}+\frac{\log ^2(a)-\log (a)}{x^3}+O\left(\frac{1}{x^4}\right)$$ whcih will show the limit and how it is approached.

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  • $\begingroup$ Could you expand for me this $\Exp(\frac{1}{1+x})$? Because I have some problems with this thing, I get: $1 + \frac{1}{x+1} + \frac{1}{2(x+1)^2} +...$ but how can I use it later? Because I will probably not able to subtract those terms. $\endgroup$ – FNTE Jan 17 '17 at 9:32
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The best idea is probably to expand in power series (if you are allowed to do that). As Wolfram Alpha confirms, $$ 2017^\frac{1}{x} - 2017^\frac{1}{x+1} = \frac{\log (2017)}{x^2} + o(1/x^2) $$ as $x \to +\infty$, and therefore the limit equals $\log (2017)$.

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  • $\begingroup$ Why is it $\log(2017)$ because I don't see it unfortunately. $\endgroup$ – FNTE Jan 17 '17 at 9:23
  • $\begingroup$ OK, I see :) Thank you! $\endgroup$ – FNTE Jan 17 '17 at 9:24
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    $\begingroup$ I am afraid that this merely copies some WA output, which is not, as far as I am aware, what the site is about. Justifying this equivalent would be another matter, and the start of actually doing some mathematics. $\endgroup$ – Did Jan 17 '17 at 11:50
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Since $\lim_{x\to\infty}2017^{1/(x+1)}=1$, your limit is $$ \lim_{x\to\infty}2017^{1/(x+1)}x^2(2017^{1/(x^2+x)}-1) = \lim_{x\to\infty}x^2(2017^{1/(x^2+x)}-1)\\ $$ Now do $t^{-1}=x^2+x$, so $$ x^2=\frac{t+2-\sqrt{t^2+4t}}{2t} $$ and you get $$ \lim_{t\to0^+}\frac{(2017^t-1)}{t}\frac{t+2-\sqrt{t^2+4t}}{2}=\ln2017 $$

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