Question

lim x- 1 (m/(1-x^m)-n/(1-x^n))

Answer 1

\[\lim_{x\rightarrow 1} \frac{m}{1-x^m}-\frac{n}{1-x^n}\]?

Answer 2

\[\lim_{x \rightarrow 1}~~\left( \frac{m}{1-x^m} - \frac{n}{1-x^n}\right)\]

YES! :D

Answer 3

hmm

Answer 4

@AkashdeepDeb @kirbykirby can u solve that

Answer 5

lim┬(x→1)⁡(m/(1-x^m )-n/(1-x^n ))=lim┬(x→1)⁡((m(1-x^n )-n(1-x^m ))/(1-x^m )(1-x^n ) )
Put x=1+h, h→0 as x→1
=lim┬(h→0)⁡((m{1-(1+h)^n }-n{1-(〖1+h)〗^n })/{1-(1+h)^m }{1-(1+h)^n } )
=lim┬(h→0)⁡((m{1-(1+h+h^2+⋯.+h^n )}-n{1-(1+h+h^2+⋯+h^m )})/{1-(1+h+h^2+⋯+h^m )}{1-(1+h+h^2+⋯+h^n )} )
=lim┬(h→0)⁡((m{1-1-h-h^2-…-h^n }-n{1-1-h-h^2-…-h^m })/{1-1-h-h^2-…-h^m }{1-1-h-h^2-…-h^n } )
=lim┬(h→0)⁡((-mh(1+h+h^2+⋯+h^(n-1) )+nh(1+h+h^2+⋯+h^(m-1) ))/(h^2 {1+h+h^2+⋯+h^(m-1) }{1+h+h^2+⋯+h^(n-1) } ))
=lim┬(h→0)⁡((-m(1+h+h^2+⋯+h^(m-1) )+n(1+h+h^2+⋯+h^(n-1) ))/h{1+h+h^2+⋯+h^(m-1) }{1+h+h^2+⋯+h^(n-1) } )
=(-m(1+0)+n(1+0))/0(1+0)(1+0) =(-m+n)/0
Which does not exist.