Homework Help

Proof that (d/dx)(x^n) = lim ((z^n)-(x^n))/(z-x) as z->x

user profile pic

murciof | Student, College Freshman | eNoter

Posted May 13, 2013 at 12:43 AM via web

dislike 1 like

Proof that (d/dx)(x^n) = lim ((z^n)-(x^n))/(z-x) as z->x

2 Answers | Add Yours

user profile pic

oldnick | Valedictorian

Posted May 13, 2013 at 1:37 AM (Answer #1)

dislike 2 like

`d/dxx^n`   (by definition) is:

 

`lim_(h->0) ((x+h)^n-x^n)/h`   setting:   `z= x+h`  is  clear that

`z->x`  as `h->0`   .

So:    `d/dx x^n= lim_(z->0) (z^n-x^n)/(z-x)`

Indeed from geometric progressions we now that:

`z^n-x^n=(z-x) sum_(k=0)^(n-1) z^k x^(n-k)`  

So that:

`(z^n-x^n)/(z-x) =sum_(k=0)^(n-1) z^kx^(n-k-1)`

and :   `lim_(z->x) (z^n-x^n)/(z-x) = lim _(z->x) sum_(k=0)^(n-1) z^k x^(n-k-1)`

Now, between 0 and  n-1 there are n element product  `z^kx^(n-k-1)`  , that, cause continuity of function `x^n`   as `z-> x` 

`z^kx^(n-k-1) -> x^(n-1)`  

then:  `lim_(z->x) (z^n-x^n)/(z-x)= nx^(n-1)`

On the other side we know that `d/dx x^n= nx^(n-1)`

So relation holds true.

 

 

user profile pic

pramodpandey | College Teacher | Valedictorian

Posted May 13, 2013 at 3:56 AM (Answer #2)

dislike 1 like

We know that

`lim_{h->0}(f(x+h)-f(x))/h=f'(x)`

`` Let

`f(x)=x^n`

`f(z)=z^n` ,  If z-x=h  ,then

`z^n-x^n=(x+h)^n-x^n`

`=x^n+C(n,1)x^(n-1)h+C(n,2)x^(n-2)h^2+.....+C(n,n)h^n-x^n`

`=nx^(n-1)h+C(n,2)x^(n-2)h^2+........+h^n`

`Thus `

`lim_{h->0}{f(x+h)-f(x)}/h=lim_{z->x}{f(z)-f(x)}/(z-x)`

`=lim_{h->0}{h(nx^(n-1)+C(n,2)x^(n-2)h+.....+h^(n-1) )}/ h`

`=nx^(n-1)`

``

``

``

Join to answer this question

Join a community of thousands of dedicated teachers and students.

Join eNotes