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Find a limit

goody

New member
Mar 31, 2020
14
Good evening! Could anybody help me with this limit?

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

limit2.png,
but I don't know if it was helpful. Thank you!
 

Cbarker1

Active member
Jan 8, 2013
236
What is the variable in the limit symbol?

Cbarker1
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
658
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 ...

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

zoeyw

New member
May 30, 2019
4

Country Boy

Well-known member
MHB Math Helper
Jan 30, 2018
428
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. [tex]x^n- a^n[/tex] is NOT [tex](x- a)(x^{n-1}- a^{n-1}[/tex]. If you multiply the right side of that you get [tex]x^n- a^{n-1}x^n- ax^{n-1}+ a^n[/tex]. [tex]x^n- a^n= (x- a)(x^{n-1}+ ax^{n-2}+ \cdot\cdot\cdot+ a^{n-2}x+ a^{n-1}[/tex].
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
658
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
Jan 30, 2018
428
Thank you. I am on entirely too many boards with too many different protocols!