Limit as n approaches infinity

[KTiaam]

Homework Statement

What is the limit of the given equations as n approaches infinity?

(1 + 3^n-1)/3ⁿ

 

[Dick]

Homework Statement

What is the limit of the given equations as n approaches infinity?

(1 + 3^n-1)/3ⁿ

You are supposed to show an attempt to solve it. Try it. Break it into two fractions.

 

[KTiaam]

You are supposed to show an attempt to solve it. Try it. Break it into two fractions.

lim 1/3ⁿ + lim 3^n-1/3ⁿ
n→∞ n→∞

the second equation:

you can cancel the 3^n-1
and the second equations turns out to be the same as the first so that equals:

lim 1/3ⁿ
n→∞


so in approaches e?

 

[Dick]

lim 1/3ⁿ + lim 3^n-1/3ⁿ
n→∞ n→∞

the second equation:

you can cancel the 3^n-1
and the second equations turns out to be the same as the first so that equals:

lim 1/3ⁿ
n→∞


so in approaches e?

e? No, what does 1/3^n approach? Just think about it. And 3^(n-1)/3^n isn't the same as the first term. What is it?

 

[KTiaam]

e? No, what does 1/3^n approach? Just think about it. And 3^(n-1)/3^n isn't the same as the first term. What is it?

1/3ⁿ approaches 1/ a really big number.
1 over a big number equals 0.




(3^n-1)/(3ⁿ)

thinking about this equation and plugging in values like 2, 3, 4, 5, ect.

you get

3¹/3²

3²/3³

3³/3⁴

they all equal 1/3

so 1/3?

 

[Dick]

1/3ⁿ approaches 1/ a really big number.
1 over a big number equals 0.




(3^n-1)/(3ⁿ)

thinking about this equation and plugging in values like 2, 3, 4, 5, ect.

you get

3¹/3²

3²/3³

3³/3⁴

they all equal 1/3

so 1/3?

Yes. But you didn't need to plug numbers in. 3^n=3^(n-1)*3. So 3^(n-1)/3^n=1/3.

 

[KTiaam]

Yes. But you didn't need to plug numbers in. 3^n=3^(n-1)*3. So 3^(n-1)/3^n=1/3.

so it the answer is zero?

1/3 x 0

 

[HallsofIvy]

Where did you get that "0" you are multiplying?

 

[KTiaam]

Where did you get that "0" you are multiplying?

the limit as x approaches infinity of 1/3ⁿ

 

[Dick]

so it the answer is zero?

1/3 x 0

It's 0+1/3, isn't it?