What is the sum (x - 2) + (3x + 8) if x = 2?

n200080 [17]

Looks like the given limit is

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1}$

With some simple algebra, we can rewrite

$\dfrac n{3n-1} = \dfrac13 \cdot \dfrac n{n-9} = \dfrac13 \cdot \dfrac{(n-9)+9}{n-9} = \dfrac13 \cdot \left(1 + \dfrac9{n-9}\right)$

then distribute the limit over the product,

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \lim_{n\to\infty}\left(\dfrac13\right)^{n-1} \cdot \lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}$

The first limit is 0, since 1/3ⁿ is a positive, decreasing sequence. But before claiming the overall limit is also 0, we need to show that the second limit is also finite.

For the second limit, recall the definition of the constant, e :

$\displaystyle e = \lim_{n\to\infty} \left(1+\frac1n\right)^n$

To make our limit resemble this one more closely, make a substitution; replace 9/(n - 9) with 1/m, so that

$\dfrac{9}{n-9} = \dfrac1m \implies 9m = n-9 \implies 9m+8 = n-1$

From the relation 9m = n - 9, we see that m also approaches infinity as n approaches infinity. So, the second limit is rewritten as

$\displaystyle\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8}$

Now we apply some more properties of multiplication and limits:

$\displaystyle \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m+8} = \lim_{m\to\infty}\left(1+\dfrac1m\right)^{9m} \cdot \lim_{m\to\infty}\left(1+\dfrac1m\right)^8 \\\\ = \lim_{m\to\infty}\left(\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)^m\right)^9 \cdot \left(\lim_{m\to\infty}\left(1+\dfrac1m\right)\right)^8 \\\\ = e^9 \cdot 1^8 = e^9$

So, the overall limit is indeed 0:

$\displaystyle \lim_{n\to\infty} \left(\frac n{3n-1}\right)^{n-1} = \underbrace{\lim_{n\to\infty}\left(\dfrac13\right)^{n-1}}_0 \cdot \underbrace{\lim_{n\to\infty}\left(1+\dfrac9{n-9}\right)^{n-1}}_{e^9} = \boxed{0}$

7 0

2 years ago

Mei Su had 80 coins. She gave most of them to her friends in such a way that each of her friends got at least one coin and no tw

Mama L [17]

Answer: 12 friends.

Step-by-step explanation:

the data we have is:

Mei Su had 80 coins.

She gave the coins to her friends, in such a way that every friend got a different number of coins, then we have that:

The maximum number of friends that could have coins is when:

friend 1 got 1 coin

friend 2 got 2 coins

friend 3 got 3 coins

friend N got N coins

in such a way that:

(1 + 2 + 3 + ... + N) ≥ 79

I use 79 because "she gave most of the coins", not all.

We want to find the maximum possible N.

Then let's calculate:

1 + 2 + 3 + 4 + 5 = 15

15 + 6 + 7 + 8 + 9 + 10 = 55

now we are close, lets add by one number:

55 + 11 = 66

66 + 12 = 78

now, we can not add more because we will have a number larger than 80.

Then we have N = 12

This means that the maximum number of friends is 12.

5 0

3 years ago

Find the length of the diagonal of a square with perimeter 32.

snow_lady [41]

Answer:

i think A. because she is 4/2

7 0

2 years ago

A rectangle has a length that is 4 times the width, w. If the perimeter of the rectangle has a perimeter of 40

Zolol [24]

Answer: 16cm

Step-by-step explanation:

l = 4w

2l + 2w = p

2(4w) + 2w = 40

8w + 2w = 40

10w = 40

w = 4 cm

l = 4w = 4(4) = 16cm

5 0

2 years ago

Use the diagram shown to find x.

marusya05 [52]

X would equal 70°.

For every triangle, all interior angles add up to 180°. This means that we can add together 42 and 68 to get 110°. Because the sum of all interior angles in a triangle must be 180°, we subtract 110 from 180 to see how much more we need, which gets us 70°.

I hope this helps!

6 0

3 years ago

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