MHB How can I continue, in order to find the limit?

[evinda]

Hello! (Wave)

How can I apply L'Hôpital's rule, in order to find this limit?$$ \lim_{n \to +\infty} \frac{n^{\sqrt n}}{2^n} $$

That's what I have tried so far:

$$ \lim_{n \to +\infty} \frac{n^{\sqrt n}}{2^n} =\lim_{n \to +\infty} \frac{e^{\sqrt{n} \ln n}}{e^{n \ln 2} }=\lim_{n \to +\infty} e^{\sqrt{n} \ln n-n \ln 2}$$

How can I continue? (Thinking)

 

[Siron]

Do you really need to use L'Hopital's rule?

In fact, I tried some ways to solve the limit only using l'Hopital's rule and standard limit properties but none of them led me to an answer. Maybe I missed something elementary. If not you could try to look at the behaviour of the function $n^{\sqrt{n}}$ and $2^n$ and see how they relate to each other for a large $n$.

 

[I like Serena]

Hello! (Wave)

How can I apply L'Hôpital's rule, in order to find this limit?$$ \lim_{n \to +\infty} \frac{n^{\sqrt n}}{2^n} $$

That's what I have tried so far:

$$ \lim_{n \to +\infty} \frac{n^{\sqrt n}}{2^n} =\lim_{n \to +\infty} \frac{e^{\sqrt{n} \ln n}}{e^{n \ln 2} }=\lim_{n \to +\infty} e^{\sqrt{n} \ln n-n \ln 2}$$

How can I continue? (Thinking)

Hi again! (Wave)

What is:
$$\lim_{n \to \infty} \sqrt{n} \ln n-n \ln 2$$
(Wondering)

 

[I like Serena]

Hey Siron! :)

If not you could try to look at the behaviour of the function $n^{\sqrt{n}}$ and $2^n$ and see how they relate to each other for a large $n$.

I believe that's where L'Hôpital comes in.

 

[evinda]

Hi again! (Wave)

What is:
$$\lim_{n \to \infty} \sqrt{n} \ln n-n \ln 2$$
(Wondering)

I tried to multiply by the conjugate expression, and I got the following:

$$\lim_{n \to +\infty} \frac{(\sqrt{n} \ln n-n \ln 2) \cdot (\sqrt{n} \ln n+n \ln 2)}{\sqrt{n} \ln n+n \ln 2}=\lim_{n \to +\infty} \frac{n \ln^2 n-n^2 \ln^2 2}{\sqrt{n} \ln n+ n \ln 2}=\lim_{n \to +\infty} \frac{n^2 \left (\frac{\ln^2 n}{n}-\ln^2 2 \right )}{n \left ( \frac{\ln n}{\sqrt{n}}-\ln 2\right )}=\lim_{n \to +\infty}\frac{n \left (\frac{\ln^2 n}{n}-\ln^2 2 \right )}{\left ( \frac{\ln n}{\sqrt{n}}-\ln 2\right )}=+\infty $$

So, the limit is equal to $+\infty$.

Is it right or have I done something wrong? (Thinking)

 

[I like Serena]

I tried to multiply by the conjugate expression, and I got the following:

$$\lim_{n \to +\infty} \frac{(\sqrt{n} \ln n-n \ln 2) \cdot (\sqrt{n} \ln n+n \ln 2)}{\sqrt{n} \ln n+n \ln 2}=\lim_{n \to +\infty} \frac{n \ln^2 n-n^2 \ln^2 2}{\sqrt{n} \ln n+ n \ln 2}=\lim_{n \to +\infty} \frac{n^2 \left (\frac{\ln^2 n}{n}-\ln^2 2 \right )}{n \left ( \frac{\ln n}{\sqrt{n}}-\ln 2\right )}=\lim_{n \to +\infty}\frac{n \left (\frac{\ln^2 n}{n}-\ln^2 2 \right )}{\left ( \frac{\ln n}{\sqrt{n}}-\ln 2\right )}=+\infty $$

So, the limit is equal to $+\infty$.

Is it right or have I done something wrong? (Thinking)

How did you find the limit in the last step? (Wondering)
It doesn't look right.Actually, I was thinking of a slightly different approach.
How would you find out which of the 2 terms is the dominant term? (Wondering)

 

[evinda]

How did you find the limit in the last step? (Wondering)
It doesn't look right.

So, isn't it equal to $+\infty \cdot \left ( \frac{0-\ln^2 2}{0-\ln 2}\right )=+\infty \cdot \ln 2=+\infty$ ? (Worried)

Actually, I was thinking of a slightly different approach.
How would you find out which of the 2 terms is the dominant term? (Wondering)

How else could we calculate the limit? I wanted to find it, since I am asked to find which of these functions is asymptotically greater.. (Thinking)

 

[I like Serena]

So, isn't it equal to $+\infty \cdot \left ( \frac{0-\ln^2 2}{0-\ln 2}\right )=+\infty \cdot \ln 2=+\infty$ ? (Worried)

I'm afraid not.
Something is going wrong when you are substituting the $+\infty$ for $n$.

How else could we calculate the limit? I wanted to find it, since I am asked to find which of these functions is asymptotically greater.. (Thinking)

How about calculating:
$$\lim_{n\to\infty} \frac{\sqrt n \ln n}{n\ln 2}$$
to figure out which term is dominant? (Wondering)

 

[evinda]

How about calculating:
$$\lim_{n\to\infty} \frac{\sqrt n \ln n}{n\ln 2}$$
to figure out which term is dominant? (Wondering)

$$\lim_{n\to\infty} \frac{\sqrt n \ln n}{n\ln 2}=\lim_{n \to +\infty} \frac{\ln n}{\ln 2\sqrt{n}}=\lim_{n \to +\infty} \frac{\frac{1}{n}}{\frac{\ln 2}{2 \sqrt{n}}}=\lim_{n \to +\infty} \frac{2}{\ln 2 \sqrt{n}}=0$$

But how can we use this to conclude which of the functions $n^{\sqrt{n}}, 2^n$ is asymptotically greater? (Thinking)

 

[I like Serena]

$$\lim_{n\to\infty} \frac{\sqrt n \ln n}{n\ln 2}=\lim_{n \to +\infty} \frac{\ln n}{\ln 2\sqrt{n}}=\lim_{n \to +\infty} \frac{\frac{1}{n}}{\frac{\ln 2}{2 \sqrt{n}}}=\lim_{n \to +\infty} \frac{2}{\ln 2 \sqrt{n}}=0$$

Good! (Nod)

And I see you've used L'Hôpital's rule. (Wink)

But how can we use this to conclude which of the functions $n^{\sqrt{n}}, 2^n$ is asymptotically greater? (Thinking)

What does this limit mean, $\epsilon$-style? (Wondering)

 

[evinda]

Good! (Nod)

And I see you've used L'Hôpital's rule. (Wink)

(Nod)

What does this limit mean, $\epsilon$-style? (Wondering)

$$\lim_{n \to +\infty} \frac{\sqrt{n} \ln n}{n \ln2}=0$$

That means that $\forall \epsilon>0, \exists n_0$ such that $\forall n \geq n_0:$

$$|\frac{\sqrt{n} \ln n}{n \ln2}|< \epsilon \Rightarrow \sqrt{n} \ln n< \epsilon \cdot n \ln 2 \Rightarrow e^{\sqrt{n} \ln n}<e^{ \epsilon \cdot n \ln 2 }$$

For $\epsilon=1:$

$$e^{\sqrt{n} \ln n}<e^{ n \ln 2 }$$

Can we show like that, that $n^{\sqrt{n}}=O(2^n)$ ? (Thinking)

 

[I like Serena]

$$\lim_{n \to +\infty} \frac{\sqrt{n} \ln n}{n \ln2}=0$$

That means that $\forall \epsilon>0, \exists n_0$ such that $\forall n \geq n_0:$

$$|\frac{\sqrt{n} \ln n}{n \ln2}|< \epsilon \Rightarrow \sqrt{n} \ln n< \epsilon \cdot n \ln 2 \Rightarrow e^{\sqrt{n} \ln n}<e^{ \epsilon \cdot n \ln 2 }$$

For $\epsilon=1:$

$$e^{\sqrt{n} \ln n}<e^{ n \ln 2 }$$

Can we show like that, that $n^{\sqrt{n}}=O(2^n)$ ? (Thinking)

Yes, we can. (Happy)

But that doesn't really give an answer to the problem does it? (Wasntme)

 

[evinda]

Yes, we can. (Happy)

But that doesn't really give an answer to the problem does it? (Wasntme)

So, don't we conclude from that, that $2^n$ is asymptotically greater than $n^{\sqrt{n}}$ ? (Thinking)

 

[I like Serena]

So, don't we conclude from that, that $2^n$ is asymptotically greater than $n^{\sqrt{n}}$ ? (Thinking)

Couldn't they still be asymptotically equal? (Wondering)

 

[evinda]

Couldn't they still be asymptotically equal? (Wondering)

So.. what else could we do? (Thinking)

 

[I like Serena]

So.. what else could we do? (Thinking)

What do you think of the following limit now?
$$\lim_{n \to \infty} \sqrt{n} \ln n-n \ln 2$$
(Wondering)

 

[evinda]

What do you think of the following limit now?
$$\lim_{n \to \infty} \sqrt{n} \ln n-n \ln 2$$
(Wondering)

$\sqrt{n} \ln n< \epsilon \cdot n \ln 2 $

$$\lim_{n \to +\infty} \sqrt{n} \ln n-n \ln 2 < \lim_{n \to +\infty} \epsilon \cdot n \ln 2- n \ln 2=\lim_{n \to +\infty} (\epsilon-1) n \ln 2$$

This is equal to $+\infty$, when $\epsilon>1$.
It is equal to $-\infty$, when $\epsilon<1$.
And, it is equal to $0$, when $\epsilon=1$

Is this right? (Thinking) Can this help? (Worried)

 

[I like Serena]

$\sqrt{n} \ln n< \epsilon \cdot n \ln 2 $

$$\lim_{n \to +\infty} \sqrt{n} \ln n-n \ln 2 < \lim_{n \to +\infty} \epsilon \cdot n \ln 2- n \ln 2=\lim_{n \to +\infty} (\epsilon-1) n \ln 2$$

This is equal to $+\infty$, when $\epsilon>1$.
It is equal to $-\infty$, when $\epsilon<1$.
And, it is equal to $0$, when $\epsilon=1$

Is this right? (Thinking) Can this help? (Worried)

It is right and it means the the limit is smaller or equal to all of the above. (Nod)

Anyway, I suggest to pick a small $\epsilon > 0$.
You will get a stronger result when $\epsilon$ is smaller.

Can you say what the limit is now? (Wondering)

 

[evinda]

It is right and it means the the limit is smaller or equal to all of the above. (Nod)

Anyway, I suggest to pick a small $\epsilon > 0$.
You will get a stronger result when $\epsilon$ is smaller.

Can you say what the limit is now? (Wondering)

So, do we choose $0< \epsilon<1$ ? But.. how can it be that the limit is $< -\infty$ ? Does this mean that the limit is equal to $-\infty$ ? (Worried)

 

[I like Serena]

So, do we choose $0< \epsilon<1$ ?

Yes.

But.. how can it be that the limit is $< -\infty$ ?

It isn't. (Shake)Generally, suppose you have $f(n) < g(n)$ for all $n$.

Then $\lim\limits_{n\to\infty} f(n) \le \lim\limits_{n\to\infty} g(n)$.Consider for instance $\frac 1n$ and $\frac 2 n$.

What are their limits? (Wondering)

Does this mean that the limit is equal to $-\infty$ ? (Worried)

Yes. (Nod)

 

[evinda]

Yes.

Why can we suppose that $0 < \epsilon<1$ ? (Thinking)

It isn't. (Shake)Generally, suppose you have $f(n) < g(n)$ for all $n$.

Then $\lim\limits_{n\to\infty} f(n) \le \lim\limits_{n\to\infty} g(n)$.Consider for instance $\frac 1n$ and $\frac 2 n$.

What are their limits? (Wondering)

They are both equal to $0$, right? (Smile)

- - - Updated - - -

Generally, suppose you have $f(n) < g(n)$ for all $n$.

Then $\lim\limits_{n\to\infty} f(n) \le \lim\limits_{n\to\infty} g(n)$.

So, when we have $\lim_{n \to \infty} f(n)< \lim_{n \to \infty} g(n)$ can we conclude that $f(n)<g(n)$ ? (Worried)

 

[I like Serena]

Why can we suppose that $0 < \epsilon<1$ ? (Thinking)

The statement of a limit says: $\forall \epsilon > 0$.
That means that we are free to pick any $\epsilon$ we like. (Muscle)

For whatever $\epsilon$ we pick, however small, there has to be an $n_0$, however big, for which the statement holds true.

Generally, we will want to pick a small $\epsilon$, to get the strongest conclusions.

They are both equal to $0$, right? (Smile)

Yes.
So even though $\frac 1n < \frac 2n$ for all $n$, their limits are equal nonetheless. (Mmm)

 

[evinda]

The statement of a limit says: $\forall \epsilon > 0$.
That means that we are free to pick any $\epsilon$ we like. (Muscle)

For whatever $\epsilon$ we pick, however small, there has to be an $n_0$, however big, for which the statement holds true.

Generally, we will want to pick a small $\epsilon$, to get the strongest conclusions.

Yes.
So even though $\frac 1n < \frac 2n$ for all $n$, their limits are equal nonetheless. (Mmm)

A ok! (Smile)

But.. we have found that $\lim_{n \to +\infty} f(n)< \lim_{n \to +\infty} g(n)$.

Can we conclude from that that $g(n)$ is asymptotically greater that $f(n)$?

 

[I like Serena]

A ok! (Smile)

But.. we have found that $\lim_{n \to +\infty} f(n)< \lim_{n \to +\infty} g(n)$.

Can we conclude from that that $g(n)$ is asymptotically greater that $f(n)$?

We didn't.

Which $f(n)$ and $g(n)$ are you referring to? (Wondering)

We did find that:
$$\lim\limits_{n \to \infty} \sqrt n \log n - n\ln 2 = -\infty$$

 

[evinda]

We did find that:
$$\lim\limits_{n \to \infty} \sqrt n \log n - n\ln 2 = -\infty$$

We are looking for the value of $e^{\lim_{n \to \infty} \sqrt n \log n - n\ln 2}=e^{-\infty}=0$.

So, can we conclude from this that $2^n$ is asymptotically greater that $n^{\sqrt{n}}$ ? (Thinking)

 

[I like Serena]

- - - Updated - - -
So, when we have $\lim_{n \to \infty} f(n)< \lim_{n \to \infty} g(n)$ can we conclude that $f(n)<g(n)$ ? (Worried)

I'm afraid not.
What examples for $f(n)$ and $g(n)$ can you think of for which this might not be true? (Wondering)

We are looking for the value of $e^{\lim_{n \to \infty} \sqrt n \log n - n\ln 2}=e^{-\infty}=0$.

So, can we conclude from this that $2^n$ is asymptotically greater that $n^{\sqrt{n}}$ ? (Thinking)

Yes. (Dull)