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'hospital rule twice, but it doesn't work. Now I have no idea how to do it.
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Set $1/x=h$ to get $$\lim_{x\to\infty}x^2(a^{1/x}-a^{1/(x+1)})$$ $$=\lim_{h\to0^+}\dfrac{a^{h/(h+1)}\left(a^{h^2/(h+1)}-1\right)}{h^2}$$ $$=\lim_{h\to0^+}a^{h/(h+1)}\cdot\lim_{h\to0^+}\dfrac{a^{h^2/(h+1)}-1}{h^2/(h+1)}\cdot\lim_{h\to0^+}\dfrac1{h+1}$$ Now 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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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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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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