How do you find the limit of #sqrt(n^2+n) - (n)# as n approaches #oo#?

Answer 1

William Abdalla

#lim_(n->oo)sqrt(n^2+n)-n =1/2#

#lim_(n->oo)sqrt(n^2+n)-n = lim_(n->oo)((sqrt(n^2+n)-n)(sqrt(n^2+n)+n))/(sqrt(n^2+n)+n)#

#=lim_(n->oo)n/(sqrt(n^2+n)+n)#

#=lim_(n->oo)(n/n)/((sqrt(n^2+n)+n)/n)#

#=lim_(n->oo)1/(sqrt((n^2+n)/n^2)+1)#

#=lim_(n->oo)1/(sqrt(1+1/n)+1)#

#=1/(sqrt(1+0)+1)#

#=1/2#

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Answer 2

Luke Alvarez

To find the limit of sqrt(n^2+n) - n as n approaches infinity, we can simplify the expression. By multiplying the numerator and denominator by the conjugate of the expression, we can eliminate the square root:

(sqrt(n^2+n) - n) * (sqrt(n^2+n) + n) / (sqrt(n^2+n) + n)

Simplifying further, we get:

(n^2 + n - n^2) / (sqrt(n^2+n) + n)

This simplifies to:

n / (sqrt(n^2+n) + n)

As n approaches infinity, the denominator becomes dominated by the term n. Therefore, the limit of the expression is:

1 / (1 + 1) = 1/2

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.