Find the limit of this sequence as n approaches infinity (ML Boas)

[agnimusayoti]

Homework Statement

Find lim of this sequences as n approaching zero

Relevant Equations

$$\lim_{n\to\infty} (1+n^2)^\frac{1}{\ln n}$$

First I assume that
$$(1+n^2)^\frac{1}{\ln n}=\exp {\ln (1+n^2)^\frac{1}{\ln n}}
$$But,
$${\ln (1+n^2)^\frac{1}{\ln n}}={\frac{\ln (1+n^2)}{\ln n}}$$
By L'Hopital Rule, I got
$$\lim_{n\to\infty} {\frac{\ln (1+n^2)}{\ln n}}=\frac{\lim_{n\to\infty} (\frac{2n}{1+n^2})}{\lim_{n\to\infty} 1/n}$$
After simplification, by multiplying numerator by $$\frac{1/n}{1/n}$$, I got 2. But, I attempted to multiplying numerator by 1/n^2 since the highest power is 2. And I came with the result undefined because 0/0.

Can anyone explain what is my mistake? Which answer is true? Or these two answer were wrong?

Thanks a bunch, pals!

 

[skeeter]

Your statement says find the limit as n approaches zero, yet you are working with n approaching infinity.
Going with what you have posted ...

Let $$\displaystyle y = \lim_{n \to \infty} (1+n^2)^{1/\ln{n}}$$

$$\displaystyle \ln{y} = \lim_{n \to \infty} \dfrac{\ln(1+n^2)}{\ln{n}}$$

$$\displaystyle \ln{y} = \lim_{n \to \infty} \dfrac{\frac{2n}{1+n^2}}{\frac{1}{n}}$$

$$\displaystyle \ln{y} = \lim_{n \to \infty} \dfrac{2n^2}{1+n^2}$$

can you see you'll finish with $$\ln{y} = 2$$

?

 

[agnimusayoti]

UPS, i mean approaching infinity

 

[agnimusayoti]

Your statement says find the limit as n approaches zero, yet you are working with n approaching infinity.
Going with what you have posted ...

Let $$\displaystyle y = \lim_{n \to \infty} (1+n^2)^{1/\ln{n}}$$

$$\displaystyle \ln{y} = \lim_{n \to \infty} \dfrac{\ln(1+n^2)}{\ln{n}}$$

$$\displaystyle \ln{y} = \lim_{n \to \infty} \dfrac{\frac{2n}{1+n^2}}{\frac{1}{n}}$$

$$\displaystyle \ln{y} = \lim_{n \to \infty} \dfrac{2n^2}{1+n^2}$$

can you see you'll finish with $$\ln{y} = 2$$

?

Uh, so 1/n multiplied by 2/n before we take limit, right?
But why you make left side become ln y? Is that important?

 

[PeroK]

$$\lim_{n\to\infty} {\frac{\ln (1+n^2)}{\ln n}}=\frac{\lim_{n\to\infty} (\frac{2n}{1+n^2})}{\lim_{n\to\infty} 1/n}$$

First, this is inaccurate. You can't move the limits onto the numerator and denominator in cases like this, where both limts are $0$. Instead, you keep going with L'Hopital.

Using the logarithm is a general technique for solving limits involving a power. See here for example:

https://brilliant.org/wiki/limits-by-logarithms/

The idea is that since $\ln$ is a continuous one-to-one function, if we have:
$$\lim_{n \rightarrow \infty} \ln (a_n) = L$$
then
$$\lim_{n \rightarrow \infty} (a_n) = e^L$$

 

[agnimusayoti]

First, this is inaccurate. You can't move the limits onto the numerator and denominator in cases like this, where both limts are $0$. Instead, you keep going with L'Hopital.

Using the logarithm is a general technique for solving limits involving a power. See here for example:

https://brilliant.org/wiki/limits-by-logarithms/

The idea is that since $\ln$ is a continuous one-to-one function, if we have:
$$\lim_{n \rightarrow \infty} \ln (a_n) = L$$
then
$$\lim_{n \rightarrow \infty} (a_n) = e^L$$

Ah I see, so if the limit goes to form 0/0 or infinity/infinity, I should continue simplify the L'Hopital.
With the Ln, I see because e^ln (a) is a.
Thanks for explanation!

 

[SammyS]

Ah I see, so if the limit goes to form 0/0 or infinity/infinity, I should continue simplify the L'Hopital.
With the Ln, I see because e^ln (a) is a.
Thanks for explanation!

@agnimusayoti ,
Can you state L'Hopital's Rule ?

 

[agnimusayoti]

@agnimusayoti ,
Can you state L'Hopital's Rule ?

mm, I can't state the rule strictly. But, what I know is the idea of L'Hospital's rule is $lim_{n\to\infty} f(n) = f'(\infty)$ Or, I must get the derivative from the function, and find $f'(\infty)$ to find the value of f(n) as n approaching $\infty$

 

[Mark44]

But, what I know is the idea of L'Hospital's rule is $lim_{n\to\infty} f(n) = f'(\infty)$ Or, I must get the derivative from the function, and find $f'(\infty)$ to find the value of f(n) as n approaching $\infty$

NO!
$f'(\infty)$ is undefined! You can't use $\infty$ in any arithmetic expression or as the argument to a function!

 

[benorin]

Roughly speaking, if $\lim \tfrac{f}{g}$ presents that either both $\lim f$ and $\lim g$ are either both zero or both $\pm \infty$ then l'Hospital's Rule states that $\lim \tfrac{f}{g}=\lim \tfrac{f'}{g'}$. Some good reading for you I think @agnimusayoti is the wikipedia page on l'Hospital's Rule. Good Luck, try the examples on that page, run through them in your mind or on paper.

 

[zinq]

Note: L'Hôpital's rule does not say that (if numerator and denominator both approach ∞ or both approach 0) then the limit equals the (limit of the derivative of the numerator) divided by (the limit of the derivative of the denominator). That is not always true.

Instead, it says that (if numerator and denominator both approach ∞ or both approach 0) the then limit equals the limit of ((the derivative of the numerator) divided by (the derivative of the denominator)).

I hope the distinction is clear.

 

[benorin]

I had read that three times @zinq but finally the obvious dawned on me. Lol

 

[PeroK]

Note: L'Hôpital's rule does not say that (if numerator and denominator both approach ∞ or both approach 0) then the limit equals the (limit of the derivative of the numerator) divided by (the limit of the derivative of the denominator). That is not always true.

Instead, it says that (if numerator and denominator both approach ∞ or both approach 0) the then limit equals the limit of ((the derivative of the numerator) divided by (the derivative of the denominator)).

I hope the distinction is clear.

Actually, it says the converse. It says that if the limit of the quotient of the derivatives (of the denominator and numerator) is something, then the limit of the quotient of the original functions (denominator and numerator) is the same.

 

[zinq]

Yes, the statement depends on the original limit's existence. I did not address that issue but rather assumed it.

In fact, to state l'Hôpital's rule in a technically precise manner requires a few more conditions as well, but I did not want to further confuse an already confused issue.

 

[SammyS]

Yes, the statement depends on the original limit's existence. I did not address that issue but rather assumed it.

In fact, to state l'Hôpital's rule in a technically precise manner requires a few more conditions as well, but I did not want to further confuse an already confused issue.

Yes.

For this exercise, there is an extra twist, being that it asks for the limit of a sequence, not a function. This may beg the question of how to take any derivative. Of course, that can all be worked out fairly easily.

 

[ehild]

Yes.

For this exercise, there is an extra twist, being that it asks for the limit of a sequence, not a function. This may beg the question of how to take any derivative. Of course, that can all be worked out fairly easily.

l'Hôpital's rule is not needed. The fraction

$$\frac{\ln (1+n^2)}{\ln n}$$
can be written as
$$\frac{\ln\left(n^2(1+1/n^2)\right)}{\ln n} $$
Expand the logarithm and divide all terms of the numerator by the denominator
$$\frac{2\ln(n)+ln(1+1/n^2)}{\ln n} $$
The limit
$$\frac{ln(1+1/n^2)}{\ln n} $$
is easy.