Find the limit of the sequence as n tends to 1

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Homework Help Overview

The problem involves finding the limit of a sequence as n approaches 1, specifically the expression (3/(1-sqrt(x)) - (2/(1-cuberoot(x))). The discussion centers around the application of limits and L'Hôpital's rule in calculus.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the method of finding the limit by making a common denominator and expanding the expression. There are attempts to apply L'Hôpital's rule due to the indeterminate form encountered when substituting x=1. Questions arise regarding the correctness of the calculations and the necessity of applying L'Hôpital's rule multiple times.

Discussion Status

The discussion is ongoing with participants providing feedback on each other's calculations. Some guidance has been offered regarding the application of L'Hôpital's rule, and there is an exploration of the implications of errors in the initial setup. Multiple interpretations of the limit are being considered.

Contextual Notes

There is a noted confusion regarding the application of L'Hôpital's rule and the resulting forms after substitution. Participants are also addressing potential errors in the algebraic manipulation of the expression.

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Homework Statement



Find the limit of the sequence as n tends to 1

(3/(1-sqrt(x)) - (2/(1-cuberoot(x))

Homework Equations





The Attempt at a Solution



making a common denominator and expanding:
= lim [tex]\frac{1-3x^{1/3}+2x^{1/2}}{1-x^{1/3}-x^{1/2}+x^{1/6}}[/tex]

I then divided the whole thing by x^1/2 but didnt get anywhere.
Any help would be v much appreciated.
thank you
 
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sara_87 said:
= lim [tex]\frac{1-3x^{1/3}+2x^{1/2}}{1-x^{1/3}-x^{1/2}+x^{1/6}}[/tex]

Simply substituting x=1 into this gives 0/0, so your limit is in one of the forms that qualify for l'hopital's rule...
 


ok,thanks so i used L'hospital's rule and got that the limit is 0; is that right?
 


Nope, you shouldn't be getting zero...there is actually an error in your first post: [tex]x^{1/3}x^{1/2}=x^{1/3+1/2}=x^{5/6} \neq x^{1/6}[/tex]

You will have to use L'hopistal's rule twice
 


you're right
but in that case, when i use l'hospitals rule once, i get 0/-0.5 = 0
so why must i use l'hospital's rule again?
 


You should be getting 0/0 after the first time (5/6-1/3-1/2=0 not -0.5)
 


Oh my, I am so stupid! i should have known that.
Yep, i see my mistake.
L'hospital's rule twice gives limit =1/2
right?
 


Yup! :smile:
 


Thanks.
 

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