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Calculate the limit n^2/( 1 + 2 + 3 + ... + n ), x->infinity

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albimaia | Student, College Freshman | eNoter

Posted May 31, 2010 at 11:11 PM via web

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Calculate the limit n^2/( 1 + 2 + 3 + ... + n ), x->infinity

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giorgiana1976 | College Teacher | Valedictorian

Posted May 31, 2010 at 11:16 PM (Answer #1)

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To evaluate the limit of the rational function, when n tends to +inf.,we'll factorize both, numerator and denominator.

In this case, first, we'll write the sum of the denominator:

1+2+...+n = n*(n+1)/2

We'll substitute the denominator, by the result of the sum, we'll factorize  by the highest power of n, which in this case is n^2.

We'll have:

lim n^2/( 1 + 2 + 3 + ... + n )  = lim n^2/lim (1+2+3+ ... +n)

lim n^2/lim (1+2+3+ ... +n) = lim 2*n^2/lim n*(n+1)

We'll open the brackets from the denominator:

lim 2*n^2/lim n^2(1+1/n)

We'll divide by n^2:

lim 2/lim (1+1/n) = 2/(1+0)=2

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neela | High School Teacher | Valedictorian

Posted May 31, 2010 at 11:22 PM (Answer #2)

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To find Lt n^2/(1+2+...n) as x--> infinity.

Solution:

The denominator is the sum of first n natural 1+2+3+.+n =  n(n+1)/2.

So we have to find Lt n^2/(n(n+1)/2] = Lt  2n^2/(n^2+n) as n--> infinity. Since both numerator and denominator approach ifinity  which is an indeterminate form, we use L'Hospital s rule of differentiating numerator and denominator and take the limit.

 Limit of the given expression = (2n^2)'/(n^2+n)' = Lt 4n /(2n+n). Still in inf/inf form. Againg applying L'Hospoital's rule

 limit of the expression = (4n)'/(2n+1)' = 4/2 = 2

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hala718 | High School Teacher | (Level 1) Educator Emeritus

Posted June 1, 2010 at 4:18 AM (Answer #3)

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lim n^2/(1+2+3+...+n)  n--> inf

We know that 1+2+3+...+n = n(n+1)/2

==> lim 2n^2/n(n+1) = 2lim n^2/ lim n^2+n

 Let us divide by the highest power n^2

==> 2lim (n^2/n^2) / lim n^2(1+1/n)

Reduce similar:

==> 2 lim(1)/lim (1+1/n)   when n--> inf

==> 2/1+0 = 2

 

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