How can i get the limit of the sequence {[n(n+2)/n+1]-[ n^3/n^2+1]}? please explain the steps.?

Question

Thanks for your time.

Answer 1

you need to find a common denominator:

n(n+2)/(n+1) - n³/(n²+1)
= n(n+2)(n²+1)/[(n+1)(n²+1)] - n³(n+1)/[(n²+1)(n+1)]
= [n(n+2)(n²+1) - n³(n+1)]/[(n²+1)(n+1)]

If you multiply everything out and look at the leading terms you should get that the limit is 1

Answer 2

For the first part of expression, do you mean [ n(n+2) / n ] +1 ? or is it n (n+2) / (n+1) ? Because if it's the first, then n / n = 1, and you are left with just (n+2). And the same type of question for the second part of the expression.

And do you mean the limit as n goes to +infinity?

The way you wrote it, as long as n is not zero, the expression is [n + 2 +1] - [n+ 1], which is equal to +2.

Answer 3

for an alternative method, use long division on the two main terms. you'll get [n+1-1/(n+1)] - [n-n/(n^2+1)]. As n goes to infinity the two fractions wipe out leaving n + 1 - n = 1

Answer 4

The limit does not exist as on simpliflication, numerator is a polynomial of degree 4, whereas the denominator is a polynomial of degree 3. So the sequence tends to infinity which is not a recognized number.

Answer 5

{[n(n+2)/(n+1)]-[ n^3/(n^2+1)]} <-- I inserted ( ) where needed.
As n --> inf, we get inf/inf - inf/inf which is indeterminate.
Combine the two fractions getting:
(n^3 + n^2 +2n)/(n^3 +n^2 +n+1)
Now we have it in form inf/inf which can be found by using L'Hospital's rule 3 times ending up with 6/6 =1.
The limit is 1.

Answer 6

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