No L'Hospital's Rule

Using the binomial theorem seems overkill. Writing it like this \( \dfrac{1-x^m}{1-x} * \dfrac{1-x}{1-x^n} \) makes a much shorter solution apparent.

This solution goes up to rational numbers,

\(\displaystyle \lim _{ x\rightarrow 1 }{ \frac { 1-{ x }^{ m } }{ 1-{ x }^{ n } } } =\lim _{ x\rightarrow 1 }{ \frac { \left( 1-{ x }^{ \frac { 1 }{ k } } \right) \left( 1+{ x }^{ \frac { 1 }{ k } }+{ x }^{ \frac { 2 }{ k } }+\cdots +{ x }^{ km } \right) }{ \left( 1-{ x }^{ \frac { 1 }{ k } } \right) \left( 1+{ x }^{ \frac { 1 }{ k } }+{ x }^{ \frac { 2 }{ k } }+\cdots +{ x }^{ kn } \right) } } for\quad any\quad k\\ =\lim _{ x\rightarrow 1 } \frac { \left( 1+{ x }^{ \frac { 1 }{ k } }+{ x }^{ \frac { 2 }{ k } }+\cdots +{ x }^{ km } \right) }{ \left( 1+{ x }^{ \frac { 1 }{ k } }+{ x }^{ \frac { 2 }{ k } }+\cdots +{ x }^{ kn } \right) } \\ =\frac { km }{ kn } \\ =\frac { m }{ n } \)

let S=limit as x tends to 1, (1-x^m)/(1-x^n) =limit as p tends to zero (1-e^pm)/(1-e^pn) ,where e^p=x.
applying expansion for e^pm and e^pn in numerator and denominator, S=lim as p tends to zero, (1-(1 + pm/1 + pm^2/2x1 + .........))/ (1-(1+ pn/1 + pn/2x1 +......)) , 1- 1 =0, in numerator and denominator , then taking ' m 'and 'n' outside from numerator and denominator , and substituting p=0 for rest of values, we get m/n. :) .pls correct me frnds, if there is any mistakes in my proof.

well one can easily use taylor series, but a taylor series expansion is in a way equivalent to L'hospital rule when you really think about it, so would you accept that ?

\(\lim _{ x-->1 }{ \frac { 1-(1+(x-1))^{ m } }{ 1-(1+(x-1))^{ n } } \quad \simeq \quad \frac { 1-(1+m(x-1)) }{ 1-(1+n(x-1)) } =\frac { m }{ n } } \)

\(Let\quad x=1+h\quad where\quad h\quad tends\quad to\quad positive\quad zero.\\ Now\quad let's\quad solve\\ \lim _{ h\rightarrow 0 }{ \frac { 1+nh }{ 1+(n-k)h } } \\ =\lim _{ h\rightarrow 0 }{ \frac { 1+nh }{ 1+nh-kh } } \\ =\lim _{ h\rightarrow 0 }{ \frac { 1+nh-kh+kh }{ 1+nh-kh } } \\ =\lim _{ h\rightarrow 0 }{ \frac { 1+nh-kh }{ 1+nh-kh } } +\frac { kh }{ 1+nh-kh } \\ =1+\lim _{ h\rightarrow 0 }{ \frac { kh }{ 1+(n-k)h } } \\ =1+\frac { 1 }{ \lim _{ h\rightarrow 0 }{ \frac { 1+(n-k)h }{ kh } } } \\ =1+\frac { 1 }{ \lim _{ h\rightarrow 0 }{ \frac { 1 }{ kh } +\frac { n-k }{ k } } } \\ Since\quad 1/kh\quad will\quad tend\quad towards\quad infinity,\quad and\quad this\quad limit\quad is\quad independent\quad of\quad other\\ "tending0s\quad or\quad tending\quad infinities",\quad \frac { n-k }{ k } \quad (a\quad finite\quad value)\quad can\quad be\quad neglected.\\ =1+\frac { 1 }{ \lim _{ h\rightarrow 0 }{ \frac { 1 }{ kh } } } \\ =1+kh\\ \\ So,\quad \frac { 1+nh }{ 1+(n-k)h } \sim 1+kh\quad as\quad h\quad tends\quad to\quad 0.\\ So,\quad \frac { 1+ah }{ 1+(a-n)h } X\frac { 1+(a-m)h }{ 1+ah } =\frac { 1+(a-m)h }{ 1+(a-n)h } \\ So,\quad \frac { 1+nh }{ 1+mh } =\frac { 1+(a-m)h }{ 1+(a-n)h } \\ Put\quad a=n,\\ \frac { 1+nh }{ 1+mh } =1+(n-m)h\\ If\quad n-m\quad is\quad constant,\quad then\quad \frac { 1+nh }{ 1+mh } \quad is\quad constant.\\ So\quad \frac { 1+(m+k)h }{ 1+mh } =\frac { 1+(m+2k)h }{ 1+(m+k)h } .\quad Put\quad m=0\\ 1+kh=\frac { 1+2kh }{ 1+kh } \\ { (1+kh) }^{ 2 }=1+2kh\\ Put\quad k=1\\ \\ We\quad can\quad easily\quad show\quad these\\ { (1+h) }^{ 2 }=1+2h\\ { (1+h) }^{ 4 }=1+4h\\ \\ { (1+(1/2)h) }^{ 2 }=1+h\\ So,\quad { (1+h) }^{ 1/2 }=1+1/2h\\ \^ \^ \quad I\quad did\quad this\quad just\quad to\quad show\quad it\quad works\quad for\quad fractions\quad also.\\ \\ So,\quad { x }^{ n }=1+nh\quad as\quad x\quad tends\quad to\quad 1\quad and\quad h\quad tends\quad to\quad 0.\\ So\quad \frac { 1-{ x }^{ m } }{ 1-{ x }^{ n } } =\frac { 1-(1+mh) }{ 1-(1+nh) } =\boxed { \frac { m }{ n } } \)