Limit as n approaches infinity

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Homework Statement



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

(1 + 3n-1)/3n
 

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  • #2
Dick
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Homework Statement



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

(1 + 3n-1)/3n
You are supposed to show an attempt to solve it. Try it. Break it into two fractions.
 
  • #3
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You are supposed to show an attempt to solve it. Try it. Break it into two fractions.
lim 1/3n + lim 3n-1/3n
n→∞ n→∞

the second equation:

you can cancel the 3n-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?
 
  • #4
Dick
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lim 1/3n + lim 3n-1/3n
n→∞ n→∞

the second equation:

you can cancel the 3n-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?
 
  • #5
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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?
1/3n approaches 1/ a really big number.
1 over a big number equals 0.




(3n-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?
 
  • #6
Dick
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1/3n approaches 1/ a really big number.
1 over a big number equals 0.




(3n-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.
 
  • #7
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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.
so it the answer is zero?

1/3 x 0
 
  • #8
HallsofIvy
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Where did you get that "0" you are multiplying?
 
  • #9
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Where did you get that "0" you are multiplying?
the limit as x approaches infinity of 1/3n
 
  • #10
Dick
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so it the answer is zero?

1/3 x 0
It's 0+1/3, isn't it?
 

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