How do you determine whether the sequence #a_n=sqrt(n^2+n)-n# converges, if so how do you find the limit?

Answer 1

See below.

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

but

# n/(sqrt(n^2+n)+n)=n/(n(sqrt(1+1/n)+1))=1/(sqrt(1+1/n)+1)#

so

#lim_(n->oo)a_n=lim_(n->oo)1/(sqrt(1+1/n)+1) = 1/2#
so the sequence is convergent and converges to #1/2#
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Answer 2

To determine whether the sequence (a_n = \sqrt{n^2 + n} - n) converges, we can analyze its behavior as (n) approaches infinity.

We can rewrite the sequence as (a_n = \frac{\sqrt{n^2 + n} - n(\sqrt{n^2 + n} + n)}{\sqrt{n^2 + n} + n}).

As (n) approaches infinity, we can simplify the expression in the numerator and denominator using the properties of limits.

After simplification, we get (a_n = \frac{1}{\sqrt{1 + \frac{1}{n}} + 1}).

Now, we can find the limit of (a_n) as (n) approaches infinity.

Taking the limit, we have (\lim_{n \to \infty} a_n = \lim_{n \to \infty} \frac{1}{\sqrt{1 + \frac{1}{n}} + 1}).

By applying limit properties, we find that the limit equals (\frac{1}{\sqrt{1 + 0} + 1} = \frac{1}{1 + 1} = \frac{1}{2}).

Therefore, the sequence (a_n = \sqrt{n^2 + n} - n) converges, and its limit is (\frac{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.

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.

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.

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.

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