Question

lim x -->1 cuberoot of x-1/ sqrt of x-1

Answer 1

well what do you think the answer is

Answer 2

i got 2...but I am not sure

Answer 3

\(\color{#000000}{\displaystyle\lim_{x \rightarrow ~1} \frac{\sqrt[3]{x-1}}{\sqrt{x-1}} }\)

like this?

Answer 4

yes :-)

Answer 5

no no no

Answer 6

it's 2

Answer 7

the cube root and square root cover only x

Answer 8

\(\color{#000000}{\displaystyle\lim_{x \rightarrow ~1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1} }\)

Answer 9

this?

Answer 10

yeah

Answer 11

Differentiate top and bottom, by L'H'S

Answer 12

So far, if you plug in x=0 into the limit, you get 0/0, therefore, you may apply L'Hospital's rule in this case,

Answer 13

I am not allowed to use l'hopital's rule

Answer 14

Interesting ...

Answer 15

yeah...I tried rationalizing but my denominator is still zero no matter what

Answer 16

I got 2 on my last try but I am still not sure 'cause i took a very sketchy approach

Answer 17

I did some rationalization, and I am getting...

\(\color{#000000 }{ \displaystyle \lim_{x\to 1}\frac{x^{5/6}-x^{1/2}+x^{1/3}-1}{x-1} }\)

not ure how that's useful

Answer 18

try a sub let x=u^6

Answer 19

\(\color{#000000 }{ \displaystyle \lim_{x\to 1}\frac{x^{1/3}-1}{(x^{1/3})^3-1} }\)

\(\color{#000000 }{ \displaystyle \lim_{x\to 1}\frac{x^{1/3}-1}{(x^{1/3}-1)(x^{2/3}+x^{1/3}+1)} }\)

Answer 20

Using,

\(\color{#000000 }{ \displaystyle a^3-b^3=(a-b)(a^2+ab+b^2) }\)

Answer 21

then cancel \(\color{#000000 }{ \displaystyle (x^{1/3}-1) }\)

Answer 22

I thought it should be x-1-1 on the numerator after rationalizing

Answer 23

you rewrote the denominator as x-1 @SolomonZelman ?

Answer 24

Also if you make the sub x=u^6
your problem becomes much easier to deal with

Answer 25

I wrote, \(\color{#000000 }{ \displaystyle x-1=(x^{1/3})^3-1}\)

Answer 26

Then, used the rule \(a^3-b^3\) ... blah blah blah

Answer 27

I am so confused!

Answer 28

What are you confused about?

Answer 29

but sqrt(x)-1 isn't x-1

Answer 30

I used that rule and after rationalizing I got...

Answer 31

oh, I see... my bad

Answer 32

I did it as if it is that, when it isn't that, but this .. anyway:)

Answer 33

\[\lim_{u \rightarrow 1}\frac{u^2-1}{u^3-1}\]
after the sub I suggested
factor a bit and cancel

Answer 34

Yeah, that works:)

Answer 35

\[((\sqrt[3]{x}-1)\div (\sqrt{x}-1)) \times ((\sqrt[3]{x})^{2}+\sqrt[3]{x}+1\div(\sqrt[3]{x})^{2}+\sqrt[3]{x}+1)\]

Answer 36

that's the rationalization right???

Answer 37

why did you divide by the denominator?

Answer 38

no...I applied...(a^3)-b^3....rule

Answer 39

The original expression is on one side...and the 'conjugate' is on the right side

Answer 40

if you are so fixed on a rationalization method over the sub method... then try...
\[\frac{x^\frac{1}{3}-1}{x^\frac{1}{2}-1} \cdot \frac{x^\frac{1}{2}+1}{x^\frac{1}{2}+1} \cdot \frac{x^\frac{2}{3}+x^{\frac{1}{3}}+1}{x^\frac{2}{3}+x^\frac{1}{3}+1} \\ =\frac{(x^\frac{1}{3}-1)(x^\frac{2}{3}+x^\frac{1}{3}+1)}{(x^\frac{1}{2}-1)(x^\frac{1}{2}+1)} \cdot \frac{x^\frac{1}{2}+1}{x^\frac{2}{3}+x^\frac{1}{3}+1}\]

Answer 41

it is like sorta rationalizing top and bottom (while really not; you are just doing it to cancel a factor )

Answer 42

that first fraction is a rational
you will see this after preforming the multiplication on top and bottom

Answer 43

honestly the sub mentioned earlier makes finding the limit easier than trying to rationalize
(my opinion)

Answer 44

Thank you guys so much for even replying in the first place...I really appreciate your help :-) @myininaya @SolomonZelman

Answer 45

Yes, indeed.

\(\color{#000000 }{ \displaystyle \lim_{x\to 1}\frac{x^{1/n}-1}{x^{1/2}-1} }\)

Will make a substitution: \(\color{#000000 }{ \displaystyle x=z^{2n} }\)
Note that as \(\color{#000000 }{ \displaystyle x \to 1 }\), \(\color{#000000 }{ \displaystyle z \to 1 }\).

\(\color{#000000 }{ \displaystyle \lim_{z\to 1}\frac{(z^{2n})^{1/n}-1}{(z^{2n})^{1/2}-1}= \lim_{z\to 1}\frac{z^2-1}{z^n-1}=\lim_{z\to 1}\frac{2z}{nz^{n-1}} =\frac{2}{n} }\)

Answer 46

That is the general derivation through L'H'S, not that it is applicable in your case, because it is not premitted as you said.

Answer 47

okay...I think i will just try the substitution method..

Answer 48

Thank you!!!:-)

Answer 49

In your example (\(x=u^6\), as myininaya has offered).
In that case, the top and bottom are factor-able ...

Answer 50

GO for it... yw

Answer 51

Okay...Thanks a bunch! @myininaya @SolomonZelman

Answer 52

Not a problem!

Answer 53

:-)