Limit of (x^3-1)/(x^1/2-1) as x approaches 1

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The limit of (x^3 - 1)/(x^(1/2) - 1) as x approaches 1 is evaluated to be 6. The discussion revolves around factorizing the numerator and denominator, specifically how to handle the zero in the denominator. One suggested method involves substituting t = x^(1/2) to simplify the limit to (t^6 - 1)/(t - 1) as t approaches 1. Participants also discuss the technique of multiplying by the conjugate to eliminate the square root in the denominator. Understanding these algebraic manipulations is key to solving the limit problem effectively.
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Homework Statement


lim x->1 (x^3-1)/(x^1/2-1)
ans:6

Homework Equations



The Attempt at a Solution


(x^1/2-1)^-1
can be converted into (plugged into wolfram):
(-x^1/2-1)/(1-x)
i want to how this done if I'm factorize out the 0 in the denominator
 
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If I'm getting it right you have

Numerator X³ -1
Denominator Sqrt(x) - 1

How can you factorize X³-1? If you know how the answer stares you right in the face.
 
ok i may have expressed myself wrongly there.
i wanted to know is how (1 / sqrt(x) - 1) can be converted into (-sqrt(x) - 1) / (1 - x)
noticed it was just multiplying denominator and numerator by its conjugate.
so yeah, should've noticed the elephant in the room.
thanks for the help anyway.
 
same way as 1/sqrt2 is sqrt2/2

you lose square root in the denominator
1*(sqrt(x) +1)/ (sqrt(x) -1)(sqrt(x) +1)
sqrt(x) +1 / (x-1) , multiply both sides of the division sign by -1 and you arrive at what you are looking for.
 
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tg43fly said:

Homework Statement


lim x->1 (x^3-1)/(x^1/2-1)
ans:6

Homework Equations



The Attempt at a Solution


(x^1/2-1)^-1
can be converted into (plugged into wolfram):
(-x^1/2-1)/(1-x)
i want to how this done if I'm factorize out the 0 in the denominator

Why not let t = x^(1/2), and so have the limit of (t^6 - 1)/(t-1) as t → 1?
 
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