Question

is the limit of 3sqrt(x) -1 / sqrt(x) -1 as x approaches 1 equal to 3/2 ?

Answer 1

I solved \[\frac{ \sqrt[3]{x} -1 }{ \sqrt{x} -1 }\]

Answer 2

i solved the limit of this equation (sorry, forgot to add lim in front)

Answer 3

so I solved it and got 3/2
is this correct?

Answer 4

Yes, that is correct :)

For future reference, if you ever want to check answers to questions like these, you can do that here: http://www.wolframalpha.com
Just type, "limit of (x^(1/3) -1)/(sqrt(x) -1) as x tends to 1"! :-)

Answer 5

thank you so much :)

Answer 6

Actually, I misread your answer... you should have got \(\frac{2}{3}\). Is that what you got?

Answer 7

hmm. I got 3/2

Answer 8

here, I will type my procedure:

Answer 9

\[\lim \frac{ x^3 -1 }{ x^2 -1 } = \lim \frac{ (x-1)(x^2+x+1) }{ (x-1)(x+1) } = \lim \frac{ x^2+x+1 }{ x+1}\]

Answer 10

= 3/2

Answer 11

Indeed, all of what you say is true :)
...but what makes you think that
\[\lim_{x\rightarrow 1} \frac{\sqrt[3]{x}-1}{\sqrt{x}-1} \text{ and } \lim_{x\rightarrow 1} \frac{x^3-1}{x^2-1}
\]are equal?

Answer 12

i just took a wild guess. I'm not surprised I was wrong there then

Answer 13

Wild guesses are extremely useful in maths, though it turns out that this time the limits are in fact, not equal.
You had the right idea though. It is possible to factor the top (as a difference of two squares) and factor the bottom, and cancel, which should leave you with a fairly clear answer.

Answer 14

hmm okay

Answer 15

I don't know what else it would equal though..

Answer 16

I think to do this question, you need to spot that the top is a difference of two squares:
\[\sqrt[3]{x}-1 = (\sqrt[6]{x}-1)(\sqrt[6]{x}+1)
\]and the bottom is the difference of two cubes. Do you know how to factorise the difference of two cubes?

Answer 17

no, I did not learn this yet. we have never used 6root(x) ever :p

Answer 18

Well, just because something hasn't explicitly been used doesn't mean it can't be :P If you want to find something that squares to a cube root, then that thing is the sixth root!

If you don't want to use this method, have you by any chance come across l'Hopital's rule? I'm guessing not, but it's worth asking...

Answer 19

no, haha. I have not come across that rule

Answer 20

Hmm... it that case, I don't really see another way to this question.

Are you familiar with indices? For example, that \(x^{\frac{1}{3}} = \sqrt[3]{x}\)?

Answer 21

yes

Answer 22

And you are familiar with the laws of indices, such as \((x^a)^b=x^{(ab)}\)?

Answer 23

yes

Answer 24

In that case, surely the statement
\[(x^{\frac{1}{6}})^2 = x^{\frac{1}{3}}\]makes sense (all I'm doing is multiplying the powers)? You might not have seen 1/6 as a power before, but it still makes sense.

Answer 25

okay ya, that makes sense

Answer 26

Then all that says is \(\sqrt[3]{x} = (\sqrt[6]{x})^2\), so you can write \(\sqrt[3]{x}\) as a square. There you have a difference of two squares at the top.

Answer 27

um, so I got lost here: \[\lim \frac{ x^{1/3} }{ (x-1)(x+1) }\]

Answer 28

The bottom does not factor to (x+1)(x-1).

\(\sqrt{x}-1\) and \(x^2-1\) are NOT the same thing! The second one does factor to (x+1)(x-1), but the first one does not.

I think the key is probably to notice that
\[\sqrt{x}-1 = x^{\frac{1}{2}} - 1 = (x^{\frac{1}{6}})^3 - (1)^3,\]which is the difference of two cubes.

Answer 29

oh ok

Answer 30

As for the top, you have
\[\sqrt[3]{x} -1 = x^{\frac{1}{3}} - 1 = (x^{\frac{1}{6}})^2 - (1)^2,\]the difference of two squares.

Answer 31

ok, yes what u are saying makes sense to me

Answer 32

Good :-) So then
\[\frac{\sqrt[3]{x} - 1}{\sqrt{x} - 1} = \frac{(x^{\frac{1}{6}}- 1)(x^{\frac{1}{6}}+ 1)}{(x^{\frac{1}{6}})^3- (1)^3}.\]I've factored the top since it's the difference of two squares. Now can you factor the difference of two cubes at the bottom?

Answer 33

um, is it:

Answer 34

\[(x^\frac{ 1 }{ 6 } -1)(x^\frac{ 1 }{ 3 } + x^ \frac{ 1 }{ 6 } +1)\]

Answer 35

I have no idea :(

Answer 36

Yup, that is correct :-)

Answer 37

okay yay :)

Answer 38

we can cross out the first parts of the numerator and denominator

Answer 39

which is x^1/6 -1

Answer 40

Exactly. Which leaves you with
\[\frac{x^{\frac{1}{6}}+1}{x^{\frac{1}{3}}+x^{\frac{1}{6}}+1}\]and the limit of each of the terms in the top and the bottom is 1.

Answer 41

Oh okay, and then that will give 2/3

Answer 42

and the only reason why my initial procedure which gave 3/2 was wrong because my identities for the cube roots and square roots were not correct, right?

Answer 43

I guess so. You incorrectly assumed that
\[\frac{\sqrt[3]{x} -1}{\sqrt{x} -1} = \frac{x^3-1}{x^2-1}.\]

Answer 44

yes, alright. thank you so much :)

Answer 45

You're welcome :)