These Long Years

$L = {\lim _{ x\rightarrow 1 }{ \left( \frac { 2018 }{ 1-{ x }^{ 2018 } } -\frac { 1998 }{ 1-{ x }^{ 1998 } } \right) } } = \dfrac{2018\,(1-x^{1998} )- 1998\,(1-x^{2018})}{ \,(1-x^{2018}\,(1-x^{1998})}$ Simplifying the the function we get $L =\lim_{x\to 1} \left(\dfrac{2018 - 2018\cdot x^{1998} -1998 +1998x^{2018}}{1 -x^{1998} -x^{2018} + x^{2018+1998}}\right)$ if we input $x=1$ them limit of the function $L= \dfrac{0}{0}$ . Hence , we can apply L-hopital rule. $L = \lim_{x\to 1}\left( \dfrac{0 - 2018\cdot 1998x^{1997} -0+ 1998\cdot 2018 x^{2017}}{0- 1998\cdot x^{1997} -2018\cdot x^{2017}+ \,(2018+1998)x^{2018+1998-1}}\right)$ Again if we input $x=1$ then limit is still $\dfrac{0}{0}$ so we need to apply L-hopital rule once more. $L = \lim_{x\to 1} \left(\dfrac{-2018\cdot 1998\cdot 1997x^{1996}+ 1998\cdot 2018\cdot 2017x^{2016} }{-1998\cdot 1997 x^{1996} -2018\cdot 2017x^{2016}+\,(2018+1998)\,(2018+1997)x^{2018+1998-2}}\right) \\ \therefore L = (\dfrac{-2018\cdot 1998\cdot 1997+ 1998\cdot 2018\cdot 2017 }{-1998\cdot 1997 -2018\cdot 2017+\,(2018+1998)\,(2018+1997)} = \dfrac{2018\cdot 1998\,(2017-1997)}{2\cdot 2018\cdot 1998 } =\boxed{10}$ Therefore, the limit of the function is $\boxed{10}$ .

The limit $\large {\lim _{ x\rightarrow 1 }{ \left( \frac { m }{ 1-{ x }^{m } } -\frac { n }{ 1-{ x }^{n } } \right) } } =\ ?$ can be generalized to $\frac{m-n}{2}$ .