Determine the limit of the convergent sequence

[ArthurRead]

Homework Statement

Determine the limit of the convergent sequence:
$a_n$ =$"3/n" ^ "1/n"$
http://www.wolframalpha.com/input/?i=lim+as+x+approaches+infinity++%283%2Fx%29^%281%2Fx%29

Homework Equations

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3. The Attempt at a Solution [/b

So I tried to get this series to look like 0/0 or ∞/∞ so I could use the L'Hopital rule.
I took the Ln of both sides. I got (ln(3/n))/n. As n approaches ∞, the numerator is -∞ and the denominator is ∞. Can
I use the L'Hopital's rule now? THANKS!

 

[STEMucator]

Given the way you've presented the sequence, this rule will be extremely useful to you :

(a/b)^c = a^c / b^c

Big hint : Remember that 1 over something really really big... aka ∞ tends to 0.

Example :

lim x->∞ 1/x = 0

 

[ArthurRead]

Thank you,
So I get (3^(1/n)) / (n^(1/n))
If n approaches ∞, the the numerator is 1 because the exponent approaches 0.
So would it be sufficient for me to apply the same mentality to denominator and say
that it approaches 1? That would give me 1 for the limit of the sequence which is the correct
answer.

 

[STEMucator]

Thank you,
So I get (3^(1/n)) / (n^(1/n))
If n approaches ∞, the the numerator is 1 because the exponent approaches 0.
So would it be sufficient for me to apply the same mentality to denominator and say
that it approaches 1? That would give me 1 for the limit of the sequence which is the correct
answer.

Yup :) good job. Sometimes the most basic rules can save you the most headaches.

 

[I like Serena]

Homework Statement

Determine the limit of the convergent sequence:
a(sub n)=(3/n)^(1/n)
http://www.wolframalpha.com/input/?i=lim+as+x+approaches+infinity++%283%2Fx%29^%281%2Fx%29

Homework Equations

--

3. The Attempt at a Solution [/b

So I tried to get this series to look like 0/0 or ∞/∞ so I could use the L'Hopital rule.
I took the Ln of both sides. I got (ln(3/n))/n. As n approaches ∞, the numerator is -∞ and the denominator is ∞.

Welcome to PF, ArthurRead!

Can I use the L'Hopital's rule now? THANKS!

Yes...

 

[I like Serena]

And no, you cannot just say that $n^{1/n}$ approaches 1.

Consider that $n^{1/2}$ approaches infinity.
And that $2^{1/n}$ approaches 1.
$n^{1/n}$ could approach anything from 1 up (although it does approach 1 ;)).

To illustrate this: $(1+{1 \over n})^n$ approaches the mathematical constant e.

 

[ArthurRead]

I am sorry "I Like Serena", I don't understand what you mean by "n1/n could approach anything from 1 up". What ever value n has, you will be taking it to the power of 0. So it has to approach 1. I know I am wrong but I don't understand how.

 

[I like Serena]

You might also say that whatever value (1/n) has, you will be taking infinity to the power of this number.
So then it would have to approach infinity.

 

[ArthurRead]

That makes sense. So am I better off going with my original approach of using L'Hops rule after taking the natural log of both sides?

 

[I like Serena]

Yes.

 

[ArthurRead]

I am not getting the right answer. (ln(3/n))/n: the derivative of the numerator is (-1/n) and the derivative of the denominator is 1. Which gives me lim = 0. Did I differentiate correctly?

 

[I like Serena]

Actually, that looks good. :)

Remember which limit you approached?

 

[ArthurRead]

Ah, that was the limit of natural log of the sequence. Since ln(a sub n) = 0, (a sub n) = e^0=1. Thank you so much for guiding me through this. I feel so much better.

by the way, is there an easier way to write (a sub n)?...I'm new to this.

 

[I like Serena]

Good!

Try:

##a_n##

That comes out as $a_n$.

Similarly

$$\lim_{n \to \infty} a_n$$

comes out as:
$$\lim_{n \to \infty} a_n$$

 

[ArthurRead]

Got it, Thanks again