Solve a limit with a nth root, with n -> infinity

[coltson]

Homework Statement

Solve the
##\lim_{n \rightarrow +\infty} \sqrt [n] \frac {n²+1} {n⁷-2} ##

3. The attempt of a solution:

First I thought about using L'Hopital's rule, but the nth root makes it useless.
Then I thought about to eliminate the root multiplying it by something that is one, but is able to modify the root's Nth power so I can start to work on it's inner.

Is that a good approach? If it is, any suggestions of the values that can be used? If it's not, what approach can I use? Thanks for the input.

 

[Orodruin]

Hint: Rewrite your expression on the form ##e^{f(n)}##, where ##f(n)## is some function of ##n##. It should be apparent what the limit of that function is and therefore how your limit behaves.

 

[Ray Vickson]

Homework Statement

Solve the
##\lim_{n \rightarrow +\infty} \sqrt [n] \frac {n²+1} {n⁷-2} ##

3. The attempt of a solution:

First I thought about using L'Hopital's rule, but the nth root makes it useless.
Then I thought about to eliminate the root multiplying it by something that is one, but is able to modify the root's Nth power so I can start to work on it's inner.

Is that a good approach? If it is, any suggestions of the values that can be used? If it's not, what approach can I use? Thanks for the input.

Often in such problems it is helpful to factor the function involved. In this case we can write
$$\frac{n^2+1}{n^7-2} = \frac{n^2}{n^7} \cdot \frac{1+\frac{1}{n^2}}{1 - \frac{2}{n^7}}= \frac{1}{n^5} \cdot \frac{1+\frac{1}{n^2}}{1 - \frac{2}{n^7}}$$
and then look at the ##n##th roots of each factor separately.

 

[coltson]

Hint: Rewrite your expression on the form ##e^{f(n)}##, where ##f(n)## is some function of ##n##. It should be apparent what the limit of that function is and therefore how your limit behaves.

I don't know how to rewrite it.

Often in such problems it is helpful to factor the function involved. In this case we can write
$$\frac{n^2+1}{n^7-2} = \frac{n^2}{n^7} \cdot \frac{1+\frac{1}{n^2}}{1 - \frac{2}{n^7}}= \frac{1}{n^5} \cdot \frac{1+\frac{1}{n^2}}{1 - \frac{2}{n^7}}$$
and then look at the ##n##th roots of each factor separately.

$$ \lim_{n \rightarrow +\infty} \sqrt [n] \frac{1}{n^5}$$ becames $$ \infty^0 $$. The rest I think follows similar paths.

 

[Ray Vickson]

I don't know how to rewrite it.
$$ \lim_{n \rightarrow +\infty} \sqrt [n] \frac{1}{n^5}$$ becames $$ \infty^0 $$. The rest I think follows similar paths.

No, absolutely not! ##\infty^0## is virtually meaningless; I can find examples where "it" equals 0, or 1, or 1,000,000 or ##+\infty.## You simply cannot avoid actually taking the limit, and in such cases, looking at the logarithm is often useful.

 

[vela]

I don't know how to rewrite it.

You could appeal to the definition of exponentiation for real numbers: ##a^b = e^{b\log a}##.

The thing that makes this problem complicated is the presence of ##n## in the exponent, right? You want to get it out of the exponent, so try taking the logarithm, as Ray suggests, and use the properties of log to make the expression easier to analyze.

$$ \lim_{n \rightarrow +\infty} \sqrt [n] \frac{1}{n^5}$$ becames $$ \infty^0.$$ The rest I think follows similar paths.

You got that backwards. The insides go to 0 and the exponent goes to infinity. Either way, it's an indeterminate form, so you need to try a different approach.

 

[coltson]

Taking the log, I have: ##
lim_{n \rightarrow +\infty} e^ {\frac 1 n * ln (\frac {n^2+1} {n^7-2}) } ##

If that is correct I end with ## e^ { 0 * ln (\frac \infty \infty) } ## with is not the answer.

 

[Ray Vickson]

Taking the log, I have: ##
lim_{n \rightarrow +\infty} e^ {\frac 1 n * ln (\frac {n^2+1} {n^7-2}) } ##

If that is correct I end with ## e^ { 0 * ln (\frac \infty \infty) } ## with is not the answer.

It is not correct: why do you continue to plug in ##n = \infty?## You just cannot do that!

Anyway, I give up.

 

[vela]

Taking the log, I have: ##
lim_{n \rightarrow +\infty} e^ {\frac 1 n * ln (\frac {n^2+1} {n^7-2}) } ##

If that is correct I end with ## e^ { 0 * ln (\frac \infty \infty) } ## with is not the answer.

##\frac \infty \infty## is an indeterminate form. You need to do a little more work to get an answer.

 

[StoneTemplePython]

##\lim_{n \rightarrow +\infty} \sqrt [n] \frac {n²+1} {n⁷-2} ##

maybe re-writing it as

##\frac{1}{n^7} \leq \frac {n²+1} {n⁷-2} \leq n²+1 \leq n^7##

would be helpful.

(assumming ##n\geq 2## )

Now, you should know that nth roots of positive numbers preserve ordering (why?).

 

[nrqed]

Taking the log, I have: ##
lim_{n \rightarrow +\infty} e^ {\frac 1 n * ln (\frac {n^2+1} {n^7-2}) } ##

If that is correct I end with ## e^ { 0 * ln (\frac \infty \infty) } ## with is not the answer.

It looks like we have to go back to basics. First, can calculate the following limit?

## \lim_{n\rightarrow \infty} \frac{1}{n} \ln n ##

 

[coltson]

It is not correct: why do you continue to plug in ##n = \infty?## You just cannot do that!

I know that (obviously). I did it as to show that I do not know how to proceed.

Anyway, I give up.

Well, thanks for nothing then.

##\frac \infty \infty## is an indeterminate form.

Yes, I know that. See my first quote in this post.

You need to do a little more work to get an answer

Well, saying that alone does not help at all. To say what kind of work might .

maybe re-writing it as

##\frac{1}{n^7} \leq \frac {n²+1} {n⁷-2} \leq n²+1 \leq n^7##

would be helpful.

(assumming ##n\geq 2## )

Now, you should know that nth roots of positive numbers preserve ordering (why?).

Well, since that for ##n>2## these inequalities are always truth, you always end having 4 numbers, that grow up from left from right. The two that are more to the right will be bigger than one . And since ## n^7## will be a bigger value than ##n^2+1##, its nth root value will also end being bigger than the other nth root value.

Now the first two values will always be smaller than one. And any nth root of number between 0 and one will always being smaller than one. So they are smaller than the values to their right.

For the remaining case I can't think in an explanation at the moment.

It looks like we have to go back to basics. First, can calculate the following limit?

Not really.

 

[stevendaryl]

Not really.

Well, as ##n \rightarrow \infty##, ##ln(n) \rightarrow \infty##. So ##\frac{ln(n)}{n}## is an indefinite form: ##\frac{\infty}{\infty}##. You can use L'Hospital's rule.