Then, find $\lim \left ( x_{n+ 1}- x_{n} \right )$

Question

Prove that for every postive integer $n$, the equation $2012^{x}\left ( x^{2}- n^{2} \right )= 1$ has a unique solution ( denoted by $x_{n}$). Then, find $$\lim \left ( x_{n+ 1}- x_{n} \right ).$$

Answer 1

If $-n\le x\le n$ then $2012^x(x^2-n^2)\le0$, so no solution.

If $x<-n$, consider $y(x)=2012^x\cdot x^2$. Then $$\frac{dy}{dx}=2012^x\left(x^2\ln2012+2x\right)=2012^x\cdot x(x\ln2012+2)$$ Now, $2012^x>0$, $x<-n<0$, and $x\ln2012+2<-n\ln2012+2\le-\ln2012+2<0$, so $$\frac{dy}{dx}>0$$ And $y(x)$ is increasing on $(-\infty,-n)$, so $$2012^x(x^2-n^2)<2012^x\cdot x^2<2012^{-n}\cdot n^2\le2012^{-1}<1$$ So no solution in this case, either.

If $x>n$ then $$\frac d{dx}\left[2012^x(x^2-n^2)-1\right]=2012^x\left((x^2-n^2)\ln2012+2x\right)>2012^x(2x)>0$$ So there can be at most one positive root $x_n$. Since $2012^n\left(n^2-n^2\right)=0$ and $$\lim_{x\rightarrow\infty}2012^x\left(x^2-n^2\right)=\infty$$ There is exactly one positive root.

We have $$0\le x_n^2-n^2=\frac1{2012^{x_n}}<\frac1{2012^n}$$ So $$\lim_{n\rightarrow\infty}\left(x_n^2-n^2\right)=0$$ By the squeeze theorem. Since $x_n>n$, $x_n+n>2n>1$, so $$\lim_{n\rightarrow\infty}\left(x_n-n\right)=0$$ Then $$\lim_{n\rightarrow\infty}\left(x_{n+1}-x_n\right)=\lim_{n\rightarrow\infty}\left((x_{n+1}-n-1)-(x_n-n)+1\right)=0-0+1=1$$