Question

Could anyone please help and explain to me how to evaluate the limit of: lim x->b (x^3-b^3)/(rootx - rootb)

Answer 1

\[\lim_{x \rightarrow b}\frac{ x ^{3}-b ^{3} }{ \sqrt{x}-\sqrt{b} }\] in the end I had gotten
\[2\sqrt{b}\times3b ^{2}\] but I'm not sure if that's right :/

Answer 2

How did you get that?

Answer 3

You know about conjugates?

Answer 4

I would start by factoring the numerator: \[
x^3-b^3=(x-b)(x^2+bx+b^2)
\]

Answer 5

I keep hearing the term "conjugates" but I don't know what it actually means, but I think I do do that...
and yea I got that, then multiplied top and bottom by (rootx + rootb)/(rootx+rootb)

Answer 6

I'll write the equation.

Answer 7

Well when you have something in the form \(a+b\) the conjugage is \(a-b\) and vice versa.

Answer 8

But usually either \(a\) or \(b\) or both is in a square root.

Answer 9

This is because:\[
(a+b)(a-b)=a^2+b^2
\]All terms are squared so there is no longer any square root.

Answer 10

oh, so I was taking the conjugate when I did the "multiplied top and bottom by (rootx + rootb)/(rootx+rootb)"

Answer 11

\[\lim_{x \rightarrow b}\frac{ x ^{3}-b ^{3} }{ \sqrt{x}-\sqrt{b} }=\lim_{x \rightarrow b}\frac{(x-b)(x^2+bx+b^2) }{ \sqrt{x}-\sqrt{b} }\frac{ \sqrt{x}+\sqrt{b} }{ \sqrt{x}+\sqrt{b} }\]

Answer 12

That simplifies to:

Answer 13

This can also be useful in imaginary numbers since \(i\) is just a another square root, in particular \(\sqrt{-1}\). Does this make sense @jigglypuff314

Answer 14

I completely failed the imaginary numbers unit in class 0.o but I understand what conjugate means now, thanks

Answer 15

\[\lim_{x \rightarrow b}\frac{(x-b)(x^2+bx+b^2) (\sqrt{x}+\sqrt{b})}{(x-b)}=\lim_{x \rightarrow b}(x^2+bx+b^2) (\sqrt{x}+\sqrt{b})\]

Answer 16

yes, then I plugged in b for x

Answer 17

So\[\lim_{x \rightarrow b}(x^2+bx+b^2) (\sqrt{x}+\sqrt{b})=3b ^{2}2\sqrt{b}=6b ^{5/2}\]

Answer 18

I hope it helped :)

Answer 19

i got lost between the last two steps. I had forgotten what rule allowed you to do that..

Answer 20

Basically you can multiply the 2 and the 3 to get a six. And then there is arule whiich says:\[x ^{a}x ^{b}=x ^{a+b}\]

Answer 21

okay thank you :) and may I ask another @ivancsc1996 ? does
\[\lim_{x \rightarrow 2^{-}}\frac{ 8-x ^{3} }{ \left| x ^{4}-16 \right| } \rightarrow \frac{ 3 }{ 8 }\]

Answer 22

Im sorry but Im not sure how to handle the absolute values.

Answer 23

if it factors out to -(x-2)(x^3+2x^2+4x+8)

Answer 24

the absolute value part i mean

Answer 25

yes it is \(\dfrac{3}{8}\)