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Find the limit of the sequence as n tends to 1

  • Thread starter sara_87
  • Start date
763
0
1. Homework Statement

Find the limit of the sequence as n tends to 1

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

2. Homework Equations



3. 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
1. Homework Statement



2. Homework Equations



3. The Attempt at a Solution
 

Answers and Replies

gabbagabbahey
Homework Helper
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5,001
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Re: limit

= 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...
 
763
0
Re: limit

ok,thanks so i used L'hospital's rule and got that the limit is 0; is that right?
 
gabbagabbahey
Homework Helper
Gold Member
5,001
6
Re: limit

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
 
763
0
Re: limit

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?
 
gabbagabbahey
Homework Helper
Gold Member
5,001
6
Re: limit

You should be getting 0/0 after the first time (5/6-1/3-1/2=0 not -0.5)
 
763
0
Re: limit

Oh my, im so stupid!!! i should have known that.
Yep, i see my mistake.
L'hospital's rule twice gives limit =1/2
right?
 
gabbagabbahey
Homework Helper
Gold Member
5,001
6
Re: limit

Yup! :smile:
 
763
0
Re: limit

Thanks.
 

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