No L'Hospital's Rule

Can you prove that $\displaystyle \lim _{ x\rightarrow 1 }{ \frac { 1-{ x }^{ m } }{ 1-{ x }^{ n } } } =\frac { m }{ n }$ without using L'Hospital's rule for counting number m and n?

Hint: I did it by using the binomial theorem.
Bonus points if you can prove it for real number m and n, because my proof can't do that and I'd be interested to see it done.

Note by Cole Wyeth
6 years, 5 months ago

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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.

- 6 years, 5 months ago

How about the extension to real numbers?

It seems like L'Hopital needs to be used.

Staff - 6 years, 5 months ago

This solution goes up to rational numbers,

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

- 5 years, 3 months ago

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.

- 6 years, 5 months ago

@ Archit Boobna - " Since 1/kh will tend towards infinity..." You need grouping symbols around kh when it is written out horizontally, because of the Order of Operations:

1/(kh)

"We can easily show these (1+h)^2 = 1 + 2h (1+h)^4 = 1 + 4h (1+(1/2)h)^2 = 1 + h"

The above are not true. You cannot use equalities. You can use the equivalents of "leads to."

- 3 years ago

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 ?

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

- 6 years, 5 months ago

I suppose most of math is pretty much equivalent- All this stuff is derived from limits, right? I think that counts.

- 6 years, 5 months ago

- 6 years, 5 months ago

Seems hardcore, but I'm a little confused by how you defined your variables at the beginning of the proof. Am I missing something?

- 6 years, 5 months ago

I am sorry, this is a bit confusing because I used "m" and "n" in the beginning also and the end also, but they have completely different roles

- 6 years, 5 months ago

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