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

[Lo.Lee.Ta.]

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.

 

[Ray Vickson]

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.

 

[Lo.Lee.Ta.]

AGH! Taking logarithms seems harder than it should be...

Just setting up a log would look like this, right? log_1+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]

AGH! Taking logarithms seems harder than it should be...

Just setting up a log would look like this, right? log_1+nx = 1/n

Not a very convenient log base. Try log_e (AKA ln).

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

But isn't there an intuitive way to figure out the right answer?

 

[Premat]

AGH! Taking logarithms seems harder than it should be...

Just setting up a log would look like this, right? log_1+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]

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.

 

[Lo.Lee.Ta.]

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]

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?

 

[Lo.Lee.Ta.]

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]

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.