# Find a limit

#### goody

##### New member
Good evening! Could anybody help me with this limit?

I have a problem when there are two variables. The only thing I did was that:

,
but I don't know if it was helpful. Thank you!

#### Cbarker1

##### Active member
What is the variable in the limit symbol?

Cbarker1

#### skeeter

##### Well-known member
MHB Math Helper
Good evening! Could anybody help me with this limit?

View attachment 10126
I have a problem when there are two variables. The only thing I did was that:

View attachment 10129,
but I don't know if it was helpful. Thank you!
actually, factoring out $(x-a)$ in the numerator yields ...

$\dfrac{(x-a)(x^{n-1} + x^{n-2}a + x^{n-3}a^2 + ... + xa^{n-2} + a^{n-1} - na^{n-1})}{(x-a)^2}$

I played with the algebra to see if I could get the other $(x-a)$ factor in the denominator to divide out ... no joy, yet.

So, starting again ...

$\displaystyle \lim_{x \to a} \dfrac{(x^n-a^n) - na^{n-1}(x-a)}{(x-a)^2}$

using L'Hopital ...

$\displaystyle \lim_{x \to a} \dfrac{nx^{n-1} - na^{n-1}}{2(x-a)}$

$\displaystyle \lim_{x \to a} \dfrac{n(x^{n-1} - a^{n-1})}{2(x-a)}$

L'Hopital again ...

$\displaystyle \lim_{x \to a} \dfrac{n(n-1)(x^{n-2})}{2} = \dfrac{n(n-1)(a^{n-2})}{2}$

#### Country Boy

##### Well-known member
MHB Math Helper
Good evening! Could anybody help me with this limit?

View attachment 10126
I have a problem when there are two variables.
No, there aren't. There are three symbols, x, a, and n. But only x is changing, it is going to a. a and n are constants. Only x is a variable.
The only thing I did was that:
View attachment 10129,
but I don't know if it was helpful. Thank you!
No. $$x^n- a^n$$ is NOT $$(x- a)(x^{n-1}- a^{n-1}$$. If you multiply the right side of that you get $$x^n- a^{n-1}x^n- ax^{n-1}+ a^n$$. $$x^n- a^n= (x- a)(x^{n-1}+ ax^{n-2}+ \cdot\cdot\cdot+ a^{n-2}x+ a^{n-1}$$.

#### skeeter

##### Well-known member
MHB Math Helper
For Country Boy, "tex" tags don't work. Dollar signs and MATH tags still work.

Quoting the last part of your post ...

No. $x^n- a^n$ is NOT $(x- a)(x^{n-1}- a^{n-1})$.
If you multiply the right side of that you get $x^n- a^{n-1}x^n- ax^{n-1}+ a^n$.

$x^n- a^n= (x- a)(x^{n-1}+ ax^{n-2}+ \cdot\cdot\cdot+ a^{n-2}x+ a^{n-1})$.

#### Country Boy

##### Well-known member
MHB Math Helper
Thank you. I am on entirely too many boards with too many different protocols!