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

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### 2 Answers

`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.

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)`

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