1. ## Limit fraction.

Hi,

I don't see how this leap is made.
$\lim_{x \to y} \frac{x^n-y^n}{x-y}$

equals

$\lim_{x \to y} x^{n-1} + x^{n-2}y + . . . + xy^{n-2} + y^{n-1}$

and this is even more confusing.

$=ny^{n-1}$

2. Originally Posted by craigmain
Hi,

I don't see how this leap is made.
$\lim_{x \to y} \frac{x^n-y^n}{x-y}$

equals

$\lim_{x \to y} x^{n-1} + x^{n-2}y + . . . + xy^{n-2} + y^{n-1}$

and this is even more confusing.

$=ny^{n-1}$
Your problem is not Calculus, but Algebra!

Its an algebraic identity $x^n-y^n = (x-y)(x^{n-1} + x^{n-2}y + . . . + xy^{n-2} + y^{n-1})$. You should know this basic identity....

3. Hello,

See here for a way to prove it : http://www.mathhelpforum.com/math-he...484-k-b-k.html

As for the initial limit, you can recognize a form of the difference quotient :
$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$

Just substitute $x=y+h$, with $f(t)=x^n$, in what you've been given to get the above.

4. Originally Posted by craigmain
Hi,

I don't see how this leap is made.
$\lim_{x \to y} \frac{x^n-y^n}{x-y}$

equals

$\lim_{x \to y} x^{n-1} + x^{n-2}y + . . . + xy^{n-2} + y^{n-1}$

and this is even more confusing.

$=ny^{n-1}$
It might be easier is you see some numbers for n

$
x^2-y^2 = (x-y)(x+y)
$

$x^3-y^3 = (x-y)(x^2+xy+y^2)$
$
x^4-y^4 = (x-y)(x^3 + x^2y+xy^2+y^3)
$

$
x^5-y^5 = (x-y)(x^4 + x^3y+x^2y^2+xy^3+y^4)
$

Notice the number of terms is the second $()$ and the power on the LHS (2 and 2, 3 and 3, 4 and 4, 5 and 5 etc), and that when $x \to y$ all the terms are the same.

5. Originally Posted by Moo
Hello,

See here for a way to prove it : http://www.mathhelpforum.com/math-he...484-k-b-k.html

As for the initial limit, you can recognize a form of the difference quotient :
$\lim_{h\to 0} \frac{f(x+h)-f(x)}{h}=f'(x)$

Just substitute $x=y+h$, with $f(t)=x^n$, in what you've been given to get the above.
Using a geometric series to prove that - I like that!

6. And, finally, taking x= y in $x^{n-1}+ x^{n-2}y+ \cdot\cdot\cdot+ xy^{n-2}+ y^{n-1}$ makes every term equal to $y^{n-1}$ and, since there are n terms, the sum is $ny^{n-1}$.