Find the following limit: $$ \lim_{x\to\infty} \left( (x+2017)^{1+\frac{1}{x}} \: -\: x^{1+\frac{1}{x+2017}} \right) $$
I tried to exchange the infinity to zero by $x:=\frac{1}{t}$ and then use $\lim_{t\to 0^+}t^t=1$, but it doesn't lead to anything, I can't avoid having $\infty-\infty$...
The answer is 2017, as graph showed.
In this case you cannot avoid using $a^b=\exp(b\ln(a))$.
You will need also $\frac{\ln(x)}{x}\to 0$ when $x\to+\infty$.
$f(x)=(x+2017)\times(x+2017)^\frac1x-x\times x^\frac1{x+2017}=$
$(x+2017)\times\exp(\frac{\ln(x+2017)}{x})-x\times\exp(\frac{\ln(x)}{x+2017})=$
The quantities inside the exponentials are going to $0$ so both exponentials are going to $1$.
If we simply use that : $(x+2017)\times(1+o(1))-x\times(1+o(1))=2017+o(x)\ $
then we cannot conclude because there is still a term that does not converge: $o(x)$ as $x\to+\infty$.
So we are forced to push the Taylor expansion forward, I'm afraid there is no shortcut this time.
$\exp\left(\frac{\ln(x+2017)}{x}\right)= \exp\left(\frac{\ln(x)+\frac{2017}{x}+O(\frac{1}{x^2})}{x}\right)= \exp\left(\frac{\ln(x)}{x}+O(\frac{1}{x^2})\right)= 1+\frac{\ln(x)}{x}+\frac{\ln(x)^2}{2x^2}+O(\frac{1}{x^2})$
$\exp\left(\frac{\ln(x)}{x+2017}\right)= \exp\left(\frac{\ln(x)}{x}-\frac{2017\ln(x)}{x^2}+O(\frac{1}{x^2})\right)= 1+\frac{\ln(x)}{x}-\frac{2017\ln(x)}{x^2}+\frac{\ln(x)^2}{2x^2}+O(\frac{1}{x^2})$
And then report in $f(x)$.
$f(x)=(x+2017)\times\exp\left(\frac{\ln(x+2017)}{x}\right)-x\times\exp\left(\frac{\ln(x)}{x+2017}\right)=$
$x\times\left(0+O(\frac{1}{x^2})\right)+2017\times\left(1+\frac{\ln(x)^2}{2x^2}+O(\frac{1}{x^2})\right)=2017+O(\frac{1}{x})\to 2017$
Note: the term in $\frac{2017\ln(x)^2}{x^2}$ disappear because it is smaller than $O(\frac1x)$.
I generalize this problem in two ways.
First, replace $2017$ with $c$.
Second, make the exponents more general, so that the problem is
$\lim_{x\to\infty} \left( (x+c)^{1+\frac{a}{x}} - x^{1+\frac{b}{x+c}} \right) $.
It turns out that the result is $(a-b)\ln(x)+c+O(\frac{\ln^2(x)}{x}) $.
If $a=b$ (as in the original problem, the limit is $c$.
Otherwise the result goes to $+\infty$ if $a > b$ and goes to $-\infty$ if $a < b$.
Let's look at the parts as $x \to \infty$.
First.
$\begin{array}\\ (x+c)^{1+\frac{a}{x}} &=(x+c)(x+c)^{\frac{a}{x}}\\ &=(x+c)x^{a/x}(1+c/x)^{\frac{a}{x}}\\ &=(x+c)e^{a\ln(x)/x}e^{a\ln(1+c/x)/x}\\ &=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2})e^{(ac/x+O(1/x^2))/x}\\ &=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2})e^{ac/x^2+O(1/x^3)}\\ &=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2})(1+ac/x^2+O(1/x^3))\\ &=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2}+O(1/x^2))\\ &=(x+c)(1+\frac{a\ln(x)}{x}+O(\frac{\ln^2(x)}{x^2}))\\ &=x+c+a\ln(x)+O(\frac{\ln^2(x)}{x} )\\ \end{array} $
Second.
$\begin{array}\\ x^{1+\frac{b}{x+c}} &=xe^{b\ln(x)/(x+c)}\\ \text{and}\\ \frac{b\ln(x)}{x+c} &=\frac{b\ln(x)}{x}\frac1{1+c/x}\\ &=\frac{b\ln(x)}{x}(1-\frac{c}{x}+O(\frac1{x^2}))\\ &=\frac{b\ln(x)}{x}-\frac{bc\ln(x)}{x^2}+O(\frac{\ln(x)}{x^3})\\ \text{so}\\ e^{b\ln(x)/(x+c)} &=1+\frac{b\ln(x)}{x}-\frac{bc\ln(x)}{x^2}+O(\frac{\ln(x)}{x^3})) +O(\frac{\ln^2(x)}{x^2})\\ &=1+\frac{b\ln(x)}{x} +O(\frac{\ln^2(x)}{x^2})\\ \text{so that}\\ x^{1+\frac{b}{x+c}} &=x+b\ln(x)+O(\frac{\ln^2(x)}{x})\\ \end{array} $
Taking their difference, the $x$ cancels out and we get $(a-b)\ln(x)+c+O(\frac{\ln^2(x)}{x}) $.
Note: Wolfy confirms the difference as $2017 + 2017 \frac{1 + 2 log(x)}{x} + O((1/x)^2) $.
Making the problem more general, let us consider $$(x+a)^{1+\frac{1}{x}} \: -\: x^{1+\frac{1}{x+a}} $$ and define $$A=(x+a)^{1+\frac{1}{x}}\qquad\text{and}\qquad B=x^{1+\frac{1}{x+a}} $$ $$\log(A)=(1+\frac{1}{x})\log(x+a)=\left(1+\frac{1}{x}\right)\left(\log(x)+\log\left(1+\frac a x\right)\right)$$ Using Taylor series $$\log(A)=\log \left({x}\right)+\frac{a+\log \left({x}\right)}{x}+O\left(\frac{1}{x^2 }\right)$$ Taylor again $$A=e^{\log(A)}=x+a+\log \left({x}\right)+O\left(\frac{1}{x}\right)$$ Doing the same for $B$ $$\log(B)= \left(1+\frac{1}{x+a}\right)\log(x)=\log \left({x}\right)+\frac{\log \left({x}\right)}{x}+O\left(\frac{1}{x^2}\right)$$ $$B=e^{\log(B)}=x+\log(x)+O\left(\frac{1}{x}\right)$$ So $$A-B=a+O\left(\frac{1}{x}\right)$$
Using one more term in the expansions, you would obtain $$A-B=a+a\frac{1+2 \log \left({x}\right)}{x}+O\left(\frac{1}{x^2}\right)$$
The number $2017$ just reminds that the question appeared in a test conducted in year $2017$. To simplify typing it is best to use a generic symbol $k$ instead of $2017$. We can then proceed as follows \begin{align} L&= \lim_{x\to\infty} (x+k) ^{1+1/x}-x^{1+1/(x+k)}\notag\\ &=\lim_{x\to\infty}(x+k) (x+k) ^{1/x}-x\cdot x^{1/(x+k)}\notag\\ &=\lim_{x\to\infty} k(x+k) ^{1/x}+x\{(x+k)^{1/x}-x^{1/(x+k)}\}\notag\\ &=k+\lim_{x\to\infty}x(e^{a}-e^{b})\notag\\ &=k+\lim_{x\to\infty}xe^{b}\cdot\frac{e^{a-b}-1}{a-b}\cdot(a-b)\notag\\ &=k+\lim_{x\to\infty}x(a-b)\notag\\ &=k+\lim_{x\to\infty}x\left(\frac{\log(x+k)}{x}-\frac{\log x} {x+k} \right) \notag\\ &=k+\lim_{x\to\infty} x\log x\left(\frac{1}{x}-\frac{1}{x+k}\right)+\log(1+(k/x))\notag\\ &=k+\lim_{x\to\infty}k\cdot\frac{\log x} {x+k} +0\notag\\ &=k+k\cdot 0=k\notag \end{align} and hence the desired answer is $k=2017$. Note that $a, b$ tend to $0$ as $x\to\infty$ and we have used the following standard limits $$\lim_{t\to 0}\frac{e^{t}-1}{t}=1,\,\lim_{x\to\infty}\frac{\log x} {x} =0$$
