What's the limit?

Question

The limit of x goes to positive infinity of
1/4*(1/3+1/8-1/(n+1)*(n+4) )
Thanks

Answer 1

Let's look at the factor in parentheses. The first two terms, 1/3 and 1/8, are constant numbers. They can just be added to the limit. The last term is the most interesting one:

1/[(n+1)*(n+4)] n --> +infinity

(I assume that both factors are part of the denominator.)

As n becomes larger, the denominator (n+1)*(n+4) also becomes larger, even infinite, and therefore 1/[...] has the limit zero.

Adding the 1/3 and 1/8, we conclude that the limit of the sum in parentheses is 1/3 + 1/8 - 0 = 11/24.

Finally, the multiplication with 1/4 can be applied to the limit, and we find

1/4 * 11/24 = 11/96.

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If you meant the last term to be [1/(n+1)] * (n+4), which is the same as (n+1)/(n+4), the limit of that term would be

lim (n+1)/(n+4) = lim [1 - 3/(n+4)] = 1

because 3/(n+4) tends to zero as n increases. In that case the result would be

(1/4) * (11/24 - 1) = (1/4) * (-13/24) = -13/96

Answer 2

I'm assuming that is a typo and you actually meant n and not x.
Also, you need to use more parentheses, so I can tell what is in the numerator and what is in the denominator.
(1/4)( 1/3 + 1/8 - (1/(n+1))(n+4)))
(1/(n+1))(n+4) goes to 1 as n goes to infinity.
The limit would be
(1/4)(1/3 + 1/8 - 1) which I'm sure you're capable of evaluating.
If n+4 is in the denominator, then 1/[(n+1)(n+4)] goes to zero. The limit would be (1/4)(1/3 + 1/8 - 0)

Answer 3

The limit is 1/4*(1/3+1/8-1/(n+1)*(n+4) ) as x is not involved.

Answer 4

lim 1/n for n--> +inf = 0
thus 1/4*(1/3+1/8-1/(n+1)*(n+4) ) --> 1/4(1/3 + 1/8)

Answer 5

lim1/4(1/3+1/8-1/n+1)(n+4))=lim1/4(11/24-0)=1/4*11/24
=11/96

Answer 6

minus infinity to plus infinity

Answer 7

THIS IS THE LIMIT!!!!!!!!!!!

Answer 8

x?

Answer 9

idk the limit does not exist