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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Sign up to join this communityI 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.
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 $\lim_{y\to0}\dfrac{a^y-1}y=\ln a$
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.
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 $\log (2017)$.
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 $$ \lim_{t\to0^+}\frac{(2017^t-1)}{t}\frac{t+2-\sqrt{t^2+4t}}{2}=\ln2017 $$