Question

limit as x approaches 0 x^8-1/x-1

Answer 1

\[\lim_{x \rightarrow 0}\frac{x^8-1}{x-1}\]

Answer 2

you can plug in x=0 right away.

Answer 3

because when you plug in x=0, you don't get any undefined values.

Answer 4

\[\lim_{x \rightarrow 0}\frac{x^8-1}{x-1}~=~\frac{(0)^8-1}{(0)-1}=\frac{-1}{-1}=1\]

Answer 5

https://www.desmos.com/calculator/q1p7e3ybo3

Answer 6

I put in 0.05 and not 0, so that it would be actually not touching the x-axis.

Answer 7

in other words,
when limits of x approach 0, the function approaches 1.

That is a point (0,1)

And this is where it looses continuity.

Answer 8

if you need more info.... tell me.
and slope is always increasing, but there are 2 concavity changes

Answer 9

are you sure you wanted \(x\to0\) ?

\(x\to1\) is a more interesting problem

Answer 10

Hmmm @Zarkon you are right!

Answer 11

yes... then you would use L'H"S .. or that would be the easiest approach.

Answer 12

L'Hopital's rule (that's the abbreviation)

Answer 13

\[\lim_{{x~} \rightarrow {~1}}~\frac{x^8-1}{x-1}=\large \lim_{{x~} \rightarrow {~1}}~\frac{\frac{d}{dx}(x^8-1)}{\frac{d}{dx}(x-1)}=\large \lim_{{x~} \rightarrow {~1}}~\frac{8x^7}{1}=\lim_{{x~} \rightarrow {~1}}~8x^7\]

Answer 14

and when you plug in 1 for x, you get 1.

Answer 15

8 actually, :P

Answer 16

ohh, man you are right.

Answer 17

LOL, how in the world did I get 1? ty for catching the mistake:)

Answer 18

Man, I had make silly mistakes too! I forgot intercepts one time lol.

Answer 19

have*

Answer 20

\[x^8-1=(x^4-1)(x^4+1)=(x^2-1)(x^2+1)(x^4+1)\]

\[=(x-1)(x+1)(x^2+1)(x^4+1)\]

Answer 21

yeah.. in my definite integrals I kept getting the wrong answer, not thinking that 1/2 - 1/4 is 1/4, not 1/2 :)

Answer 22

I like that Zarkon:)

Answer 23

Yes yes!! Zarkon just what I WAS PONDERING

Answer 24

he s a better latex typer :P

Answer 25

\[\lim_{x\to1}\frac{x^8-1}{x-1}=\lim_{x\to1}\frac{(x-1)(x+1)(x^2+1)(x^4+1)}{x-1}\]

\[=\lim_{x\to1}(x+1)(x^2+1)(x^4+1)=(1+1)(1^2+1)(1^4+1)=2\times2\times2=8\]