how do you solve the limit of n->infinite for (3^(n+1)+4^(n+1))/(3^n+4^n) ? (w/o using a calculator)?

Question

if it doesnt appear correctly, the denominator is 3^n+4^n

Answer 1

lim [ 3^(n + 1) + 4^(n + 1) ] / [ 3^n + 4^n ]
n -> infinity

One obvious option is to use L'Hospital's rule, but we can actually avoid using it.
First, split up 3^(n + 1) into 3*(3^n) and 4^(n + 1) as 4(4^n).

lim [ 3(3^n) + 4(4^n) ] / [ 3^n + 4^n ]
n -> infinity

Divide everything by 4^n

lim [ 3(3^n)/(4^n) + 4 ] / [ (3^n)/(4^n) + 1 ]
n -> infinity

Use the property (a^n)/(b^n) = (a/b)^n.

lim [ 3 (3/4)^n + 4 ] / [ (3/4)^n + 1 ]
n -> infinity

It is a fact that, for r^n, if r < 1, then
lim r^n = 0
n -> infinity
In our case, we have (3/4), which is definitely less than 1. Therefore, evaluating the limit, we get

[ 3(0) + 4 ] / [0 + 1]

[ 0 + 4 ] / [ 1 ]

(4/1)

So our final answer is 4.

Answer 2

This may be a bit casual and it is certainly not elegant.

[ 3^(n+1) + 4^(n+1) ] / [ 3^n + 4^n ]

[3 * 3^n + 4 * 4^n] / [ 3^n + 4^n]

= 3 (3^n + 4^n) / ( 3^n + 4^n) + 4^n / (3^n + 4^n)

The limit of the first term is clearly 3, leaving the second term for thought.

Hereafter, I consider only the second term, but don't forget that the first term limit is 3.

Now, I am sure there is an easier way to do this, but I don't know how. So I re-write the 2nd term in a klutzy fashion

1 / { ( 3^n + 4^n) / 4^n) }

1 / { ( 3^n / 4^n) + (4^n /4^n) }

call the denominator a + b , So the term is 1 / (a+b)

a = (3/4)^n Any number < 1 raised to big powers >> 0.

b = 1.

Therefore the denominator >> 1 as n gets very large.

1 / (a+b) >> 1 as n gets very large.

Remembering our "3" from before, the limit of the entire formula is 4.

Answer 3

This looks like a converging sequence.
first let's separate into the sum of 2 ratios
***important when you have sum in the numerator each part is over the entire denominator, you can not separate the denominator**
[3^(n+1) + 4^(n+1)]/[3^(n) + 4^(n)]
is the same as
[3^(n+1)/(3^(n) + 4^(n))] +
[4^(n+1)/(3^(n) + 4^(n))]
remember the limit of a sum is the sum of the limits so you can take each limit separately and add them together for your answer.
recall
3^(n+1)= (3^n)(3^1)= 3(3^n)
so now we're looking for the limit as n goes to infinity of
(3(3^n))/[3^(n) + 4^(n)]
multiply the numerator and denominator each by 1/(3^n) and you end up with
3/[(1 + (4^n)/(3^n))]
remembering
(4^n)/(3^n)= (4/3)^n
since, 4/3 is greater than 1, as n increases so does the ratio to infinity,
1 is very small (insignificant) compared to infinity so what we consider is
the limit as n goes to infinty of 3/n.
that will be a very small number close to 0.
there is half of what we need.
we have
0+limit n->inf [4^(n+1)/(3^(n) + 4^(n))]
we do the same process here
multiply top and bottom by 1/(4^(n))
to get
4/((3/4)^n +1)
since 3/4 is less than 1 as n increases it will get small (goes to zero)
leaving us
4/1
so the sum of the limits is
0+4
hopes this helps

Answer 4

whenever you do the limit of a problem try to break it down to as many simple steps as possible. First split it up into the numerator and the denominator. Plug in infiniti in the numerator. 3 to the infinity +1 is gonna to approach infinity. 4 to the uinfiity +1 is goin to approact infinity as well. So for the numerator you have infinity + infinity. In the denominator you have 3 to the infinity. This will approach infinity. Also you have 4 to the infinity. This approaches infinity. so for then denominator you have infinity + infiniy. So infinity + infinty is also known as infinity. You have infinity divided by infinity. So the limit is 1.

Answer 5

3^(n+1)/3^n can be written as 3^n3^1/3^n so the 3^n would cancel out, 4^(n+1)/4^n is the same as 4 so no matter what n=, the solution will be 3 times 4= 12

Answer 6

Puggy is completely right. That is the usual way to solve these problems. You generally divide by the item with the highest exponent and simplify. Then you apply the limit.