Limit n approaches infinity of (1 + n)^(1/n)

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  • #1
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1. lim (1 + n)^(1/n)
n→∞

2. I was able to figure out that the limit goes to 1 only after I substituted larger and larger values in place of n in my calculator.

Since I cannot use a calculator on my test, is there another way to know what the limit goes to?

Thanks.
 

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  • #2
Ray Vickson
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1. lim (1 + n)^(1/n)
n→∞

2. I was able to figure out that the limit goes to 1 only after I substituted larger and larger values in place of n in my calculator.

Since I cannot use a calculator on my test, is there another way to know what the limit goes to?

Thanks.
Take logarithms.
 
  • #3
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AGH! Taking logarithms seems harder than it should be...

Just setting up a log would look like this, right? log1+nx = 1/n

...Don't even know how to solve that.

But isn't there an intuitive way to figure out the right answer?
 
  • #4
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AGH! Taking logarithms seems harder than it should be...

Just setting up a log would look like this, right? log1+nx = 1/n
Not a very convenient log base. Try loge (AKA ln).
...Don't even know how to solve that.

But isn't there an intuitive way to figure out the right answer?
 
  • #5
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AGH! Taking logarithms seems harder than it should be...

Just setting up a log would look like this, right? log1+nx = 1/n

...Don't even know how to solve that.

But isn't there an intuitive way to figure out the right answer?
I would start approaching this by considering that: lim_(x→∞) for (1/x) = 0
 
  • #6
Curious3141
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1. lim (1 + n)^(1/n)
n→∞

2. I was able to figure out that the limit goes to 1 only after I substituted larger and larger values in place of n in my calculator.

Since I cannot use a calculator on my test, is there another way to know what the limit goes to?

Thanks.
Are you sure you wrote the limit correctly? There's a much more interesting limit that goes ##\displaystyle \lim_{n \to \infty} (1 + \frac{1}{n})^n## or, equivalently, ##\displaystyle \lim_{n \to 0} (1 + n)^{\frac{1}{n}}##.

As written, the limit is easily seen to be equivalent to ##\displaystyle \lim_{n \to \infty} n^{\frac{1}{n}}## since the "1" can be ignored at the limit. This limit can be resolved by taking the natural logarithm, and then using L' Hopital's Rule.
 
  • #7
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Okay, thanks, you guys! :)

So I take the ln to bring down the fraction.

ln[(1+n)1/2]

= 1/2ln(n+1) →L'Hospital's→ (1/2)(1/(n+1))(1) = 1/(2n + 2)

When I substitute infinity for n here, it seems that it goes to 0, not 1...

Am I doing something wrong here...?
 
  • #8
Dick
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Okay, thanks, you guys! :)

So I take the ln to bring down the fraction.

ln[(1+n)1/2]

= 1/2ln(n+1) →L'Hospital's→ (1/2)(1/(n+1))(1) = 1/(2n + 2)

When I substitute infinity for n here, it seems that it goes to 0, not 1...

Am I doing something wrong here...?
Where did the (1/2) power come from? And if the log goes to 0 what does the original sequence go to?
 
  • #9
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oh, man. I was looking at another problem and wrote down 1/2 instead of 1/n!

ln(1+n)1/n = ln(1+n)/n →L'Hospital's→ 1/(1+n)

1/(1+infinity) = 0

Oh, so you mean that since we used the ln, and it went to zero, then the original expression must have gone to 1.

Because ln(1) = 0

Is this right?

Thanks! :)
 
  • #10
Curious3141
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oh, man. I was looking at another problem and wrote down 1/2 instead of 1/n!

ln(1+n)1/n = ln(1+n)/n →L'Hospital's→ 1/(1+n)

1/(1+infinity) = 0

Oh, so you mean that since we used the ln, and it went to zero, then the original expression must have gone to 1.

Because ln(1) = 0

Is this right?

Thanks! :)
Pretty much right, except that I think you should use limit notation instead of writing things like 1/(1+infinity) = 0.
 

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