Question

Can someone help me with this.
Find the limit of this problem
limit as x-->0
(³√(1+2x)-1)/(x)

Answer 1

is that a cube root inclusive of -1 term in numearator, or is -1 out of numerator

Answer 2

its out of the numerator

Answer 3

i guess what i am trying to say, is does that cube root include the -1 term

Answer 4

³√(1+2x)-1

Answer 5

no it should be separate. sorry
³√(1+2x) - 1

Answer 6

limit does not exist

Answer 7

my textbook says it was 2/3.

Answer 8

well then can you please retype the question,

Answer 9

heres what I'd suggest multiply numerator and denominator by conjugate

Answer 10

(³√(1+2x)-1)/ x

Answer 11

multiply by that cube root term + 1
and in numerator you will have to differnce of two squares concept

Answer 12

which will give you 1+2x-1 and in denominator you will have (x)* ( cuberoot term +1)

Answer 13

no cancel out that x with the x in denominator

Answer 14

finally put x=0 , i.e evaluate limits as you cant simply further

Answer 15

so i got this:

(1+2x-1)/ (x (³√(1+2x)+1))

Answer 16

yes

Answer 17

cancel x from num and den

Answer 18

you said cancel out the x so, is it like this?

(1+2-1)/ (³√(1+2x)+1)

Answer 19

then,

2/ (³√(1+2x)+1)

Answer 20

but if i plug in 0, it gives me 1. i don't know what i did wrong.

Answer 21

okie, i guess you must retype the entire question

Answer 22

there seems to be a minor error in typing those brackets

Answer 23

(³√(1+2x)- 1)/ x

Answer 24

one moment, are you allowed to use L' hospital rule ?

Answer 25

cube root doesn't involve minus1

Answer 26

i haven't learned that yet

Answer 27

the correct factor to simplify this with is
\[\frac{ \sqrt[2/3]{1+2x} +1 }{\sqrt[2/3]{1+2x} +1}\]

Answer 28

because a simple look and it tells you that this qustion is of 0/0 form meaning you need to use it

Answer 29

that will give you 2/3

Answer 30

could u tell me what L-Hopsital rule is?

Answer 31

and Algebraic, how did you get that?

Answer 32

algebraic is incorrect, you cant use perfect square

Answer 33

as it 0/0 form

Answer 34

put x =0 and you will find numerator changes to 1+0-1 and denominator -=0

Answer 35

ok i get that. but how is it 2/3 then?

Answer 36

anyone?

Answer 37

hey algebraic do you know how to do this?

Answer 38

please. i need this

Answer 39

anyone still out there?

Answer 40

fml

Answer 41

5 more mintues...

Answer 42

if any1 is listening to this, i've only got a few minutes left.

Answer 43

damn i got to go... last chance any1?

Answer 44

Guess not

Answer 45

psi or algebraic, any of you still there?

Answer 46

last chace guys!!!

Answer 47

thats it screw this place.

Answer 48

You there?

Answer 49

yea

Answer 50

whelp, the way I mentioned works, but it's long. I stumbled across a simple way while playing with it however.

Answer 51

make a substitution u=2x+1
then x=(u-1)/2

and your expression looks like
\[2*\frac{ u ^{1/3} -1 }{u-1}\]

Answer 52

the denom. is a difference of cubes : u^(1/3) and 1

Answer 53

use the difference of cubes identity to expand that

Answer 54

and the numerator becomes 2 and after taking the limit the denom becomes 3:)

Answer 55

Apply L'Hospital's rule.
Using Mathematica notation below. D[f(x),x] means take the derivative of f(x) with respect to x.
\[\text{Limit}\left[\frac{D\left[\sqrt[3]{1+2x}-1,x\right]}{D[x,x]},x\to 0\right]\to \text{Limit}\left[\frac{\frac{2}{3 (1+2 x)^{2/3}}}{1},x\to 0\right] \]

Answer 56

don't think he's allowed to. thanks for bringing up l'Hopital's for the hundredth time though.

Answer 57

well thanks for your help.

Answer 58

did you follow my explanation @abc14 ?

Answer 59

yeah i think i know how to do it now

Answer 60

a^3 – b^3 = (a – b)(a^2 + ab + b^2)
a= u^(1/3) b=1

Answer 61

try it :)

Answer 62

((U^1/3)-1)(u^2/3 + u^1/3 +1)
is tthat correct algebraic

Answer 63

yep

Answer 64

that looks ugly tho

Answer 65

naw

Answer 66

numerator was 2*(u^(1/3) -1)

Answer 67

so what cancels?

Answer 68

2((U^1/3) -1)(u^2/3 + u^1/3 +1)
i don't see anything that can be

Answer 69

(u^(1/3) -1)

Answer 70

2*(u^(1/3) -1)
______________________________________
(u^1/3)-1)(u^2/3 + u^1/3 +1)

Answer 71

so 2 over (u^2/3 + u^1/3 +1)

Answer 72

lim (u^2/3 + u^1/3 +1) = 3 lol
as x->0

Answer 73

2/3

Answer 74

u^2/3 + u^1/3 = u?

Answer 75

\[2* \frac{ u ^{1/3} -1 }{ (u ^{1/3} -1)(u ^{2/3} +u ^{1/3} +1)}\]

Answer 76

\[\frac{ 2 }{u ^{2/3} +u ^{1/3} +1}\]

Answer 77

put x's back in for your u's
\[\frac{ 2 }{( (2x+1)^{2/3} +(2x+1)^{1/3} +1}\]

Answer 78

take the limit as x->0

Answer 79

ohh ok

Answer 80

thank you algebraic!

Answer 81

no problem:)