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## Homework Statement

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

(1 + 3

^{n-1})/3

^{n}

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

(1 + 3

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Dick

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

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

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

- #3

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lim 1/3You are supposed to show an attempt to solve it. Try it. Break it into two fractions.

n→∞ n→∞

the second equation:

you can cancel the 3

and the second equations turns out to be the same as the first so that equals:

lim 1/3

n→∞

so in approaches e?

- #4

Dick

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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?lim 1/3^{n}+ lim 3^{n-1}/3^{n}

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 1/3^{n}

n→∞

so in approaches e?

- #5

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1/3e? 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 over a big number equals 0.

(3

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

you get

3

3

3

they all equal 1/3

so 1/3?

- #6

Dick

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Yes. But you didn't need to plug numbers in. 3^n=3^(n-1)*3. So 3^(n-1)/3^n=1/3.1/3^{n}approaches 1/ a really big number.

1 over a big number equals 0.

(3^{n-1})/(3^{n})

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

you get

3^{1}/3^{2}

3^{2}/3^{3}

3^{3}/3^{4}

they all equal 1/3

so 1/3?

- #7

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so it the answer is zero?Yes. But you didn't need to plug numbers in. 3^n=3^(n-1)*3. So 3^(n-1)/3^n=1/3.

1/3 x 0

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HallsofIvy

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

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the limit as x approaches infinity of 1/3Where did you get that "0" you are multiplying?

- #10

Dick

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It's 0+1/3, isn't it?so it the answer is zero?

1/3 x 0

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