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

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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Ray Vickson
Homework Helper
Dearly Missed
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.

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?

Mark44
Mentor
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?

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

Curious3141
Homework Helper
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.

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...?

Dick
Homework Helper
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?

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! :)

Curious3141
Homework Helper
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.