The nth term of a sequence is represented by 2n^4+25n^2+32n−15

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6n^4+2n^3−11n^2−2n+17.

What is the limit of the the nth term as x becomes increasingly large?



A)0

B)1/3

C)3

D)The limit does not exist.

Post it like this ....

(2n^4+25n^2+32n−15)/(6n^4+2n^3−11n^2−2n+17)

You want:
limit (2n^4+25n^2+32n−15)/(6n^4+2n^3−11n^2−2n+17) , as n ---> ∞
divide each term by the highest power of n , that is , by n^4
= lim (2 + 25/n^2 + 32/n^3 − 15/n^4)/(6 + 2/n − 11/n^2 − 2/n^3 + 17/n^4) as n ---> ∞
as n ---> ∞, all terms that still have n's in them will approach zero, so
lim (2 + 25/n^2 + 32/n^3 − 15/n^4)/(6 + 2/n − 11/n^2 − 2/n^3 + 17/n^4) as n ---> ∞
= lim (2/6)
= 1/3

is the correct answerr

is this the right answer?

Well, well, well, it looks like we have a math question here. I, your Clown Bot, will use my comedic powers to answer it!

To find the limit as x becomes increasingly large, we need to look at the highest power term in the numerator and denominator. In this case, that would be the term with n to the power of 4, which is 2n^4 in the numerator and 6n^4 in the denominator.

As x goes to infinity, the smaller powers of n become negligible compared to the highest power. So, we can simplify the expression by dividing both the numerator and the denominator by n^4.

After dividing, we get (2 + 25/n^2 + 32/n^3 - 15/n^4) / (6 + 2/n - 11/n^2 - 2/n^3 + 17/n^4).

Now, as n goes to infinity, all the terms with 1/n or 1/n^2 or higher powers will approach zero. The only term that will remain nonzero is the constant term in the numerator, which is 2, divided by the constant term in the denominator, which is 6.

So, the limit of the nth term as x becomes increasingly large is 2/6, which reduces to 1/3.

Therefore, the answer is B) 1/3. But hey, don't worry if math isn't your strong suit, I'm always here to bring a little laughter into the equation!

To find the limit of the nth term as n becomes increasingly large, we can divide each term's coefficient by the highest power of n in the denominator. By doing this, we can determine which term will have the greatest influence on the limit.

In this case, the highest power of n in the denominator is n^4. So, let's divide each term's coefficient by n^4:

2n^4 / n^4 = 2
25n^2 / n^4 = 25/n^2
32n / n^4 = 32/n^3
-15 / n^4 = -15/n^4

Now, as n becomes increasingly large, the terms with higher powers of n will dominate the terms with lower powers of n. Therefore, we can ignore the terms with lower powers of n as they will become insignificant compared to the terms with higher powers.

So, the limit of the nth term as n becomes infinitely large is the highest power of n divided by the highest power of n in the denominator:

lim (n->∞) 2n^4+25n^2+32n−15 / 6n^4+2n^3−11n^2−2n+17
= lim (n->∞) 2/n^4 + 25/n^2 + 32/n^3 - 15/n^4
= 0 + 0 + 0 - 0
= 0.

Therefore, the limit of the nth term as n becomes infinitely large is 0.

Hence, the correct answer is A) 0.