Russianal Limits

$\displaystyle\lim_{x\to 0}\frac{\sqrt{x^{2}+2014^{2}}-2014}{\sqrt{x^{2}+2015^{2}}-2015}\cdot \frac{(\sqrt{x^{2}+2014^{2}}+2014)(\sqrt{x^{2}+2015^{2}}+2015)}{(\sqrt{x^{2}+2014^{2}}+2014)(\sqrt{x^{2}+2015^{2}}+2015)}$ $=\displaystyle\lim_{x\to 0}\frac{x^{2}(\sqrt{x^{2}+2015^{2}}+2015)}{x^{2}(\sqrt{x^{2}+2014^{2}}+2014)}$ $=\dfrac{2015}{2014}$

For any $m,n>0$ , we can evaluate general limit $\displaystyle \lim_{x\to0}\dfrac{\sqrt{x^2+m^2}-m}{\sqrt{x^2+n^2}-n}$ . We can apply L'Hopital's Rule here: $\begin{aligned} ~\cdots = \lim_{x\to0}\dfrac{(x^2+m^2)^{\frac12}-m}{(x^2+n^2)^{\frac12}-n}&=~\lim_{x\to0}\dfrac{\frac12(x^2+m^2)^{-\frac12}\cdot2x}{\frac12(x^2+n^2)^{-\frac12}\cdot2x} \\~\\ &=~\lim_{x\to0}\dfrac{(x^2+m^2)^{-\frac12}}{(x^2+n^2)^{-\frac12}} \\~\\ &=~\dfrac{(0^2+m^2)^{-\frac12}}{(0^2+n^2)^{-\frac12}} \\~\\ &=~\dfrac{(m^2)^{-\frac12}}{(n^2)^{-\frac12}}\\~\\ &=~\dfrac{1/m}{1/n} \\~\\ &=~\dfrac n m \end{aligned}$

Here, we have $m = 2014$ and $n=2015$ , so limit is $\dfrac{2015}{2014}$ , hence our answer is $2015 - 2014 = \boxed1$