# Math Help - Three problems

1. ## Three problems

Here are three problems I have done on other websites recently, if anyone would care to give them a go, that would be great.

Problem one

Compute

$L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2 x^2}dx$

Problem two

Let $a_1>a_2>a_3\cdots>a_n>0$

Furthermore let $p_1>p_2>p_3\cdots>p_n$

and $p_1+p_2+p_3+\cdots+{p_n}=1$

So then if $F(x)=\left(p_1a_1^x+p_2a_2^x+p_3a_3^x+\cdots+p_na_ n^x\right)^{\frac{1}{x}}$

Compute

$L=\lim_{x\to{0^+}}F(x)$

$L_=\lim_{x\to\infty}F(x)$

$L_2=\lim_{x\to-\infty}F(x)$

Problem three

Prove that if $a>1$ that

$0\leq\int_a^{\infty}\frac{\sin(x)}{x}\leq\int_0^{\ infty}\frac{\sin(x)}{x}~dx$

Note

If you have seen my solution, please don't duplicate it, for that wouldn't be fun. Plus, I always want to see other points of view!

2. I'll attempt problem 1.

$L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2 x^2}dx$

(1) I divide the numerator and denominator by n which yields
$\int_{a}^{\infty} \frac{1}{1/n+nx^2}dx$
(2) There's a formula if a>0, then
$\int \frac{1}{x^2+a^2}dx = \frac{1}{a} tan^{-1} \frac{x}{a} + C$
(3) Since $( \sqrt{n}x)^2=nx^{2}$ from (1), I use u-substitution. I let u = $\sqrt{n}x$
(4) By u substitution the integral becomes
$\int \frac{1}{1/u^2+u^2}du$ . Edit:No, this is not correct. Scratch this.
(5) Applying the formula (2) with (4) with get $tan^{-1}(nx)+C$
(6) As for the limit, $\lim_{n\to\infty} tan^{-1}(nx) = \frac{\pi}{2}$

For problem 3, I believe there's a typo.
$0\leq\int_a^{\infty}\frac{\sin(x)}{x}\leq\int_0^{\ infty}~dx\frac{\sin(x)}{x}~dx$should it be
$0 \leq \int_a^{\infty} \frac{\sin(x)}{x}~dx \leq \int_0^{\infty}\frac{\sin(x)}{x}~dx$?
I'm sure I have to use the Squeezing Principle for this. I can't seem to integrate sin(x)/x in terms of elementary functions. I'm guessing I can integrate this using the Taylor series for the sine function, but I have no idea after that.

3. Originally Posted by Paperwings
I'll attempt problem 1.

$L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2 x^2}dx$

(1) I divide the numerator and denominator by n which yields
$\int_{a}^{\infty} \frac{1}{1/n+nx^2}dx$
(2) There's a formula if a>0, then
$\int \frac{1}{x^2+a^2}dx = \frac{1}{a} tan^{-1} \frac{x}{a} + C$
(3) Since $( \sqrt{n}x)^2=nx^{2}$ from (1), I use u-substitution. I let u = $\sqrt{n}x$
(4) By u substitution the integral becomes
$\int \frac{1}{1/u^2+u^2}du$ . Edit:No, this is not correct. Scratch this.
(5) Applying the formula (2) with (4) with get $tan^{-1}(nx)+C$
(6) As for the limit, $\lim_{n\to\infty} tan^{-1}(nx) = \frac{\pi}{2}$

For problem 3, I believe there's a typo.
$0\leq\int_a^{\infty}\frac{\sin(x)}{x}\leq\int_0^{\ infty}~dx\frac{\sin(x)}{x}~dx$should it be
$0 \leq \int_a^{\infty} \frac{\sin(x)}{x}~dx \leq \int_0^{\infty}\frac{\sin(x)}{x}~dx$?
I'm sure I have to use the Squeezing Principle for this. I can't seem to integrate sin(x)/x in terms of elementary functions. I'm guessing I can integrate this using the Taylor series for the sine function, but I have no idea after that.
Your work for the first one is hard to follow (consider using white space better), but assuming you did do it partly correct (which no offense, I do no think you did), you only got a third of the answer.

As for the second one, yes thank you for the typo alert.

And $\frac{\sin(x)}{x}=\frac{\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)!}}{x}=\sum_{n=0}^{\infty}\fra c{(-1)^nx^{2n}}{(2n+1)!}$

So

$\int\frac{\sin(x)}{x}~dx=\int\sum_{n=0}^{\infty}\f rac{(-1)^nx^{2n}}{(2n+1)!}~dx=\sum_{n=0}^{\infty}\frac{(-1)^nx^{2n+1}}{(2n+1)^2(2n)!}+C$

But that is actually not how you have to do this problem, consider that it is a definite integral, and consider the ramifications of the methods able to be applied now.

4. Originally Posted by Mathstud28
Here are three problems I have done on other websites recently, if anyone would care to give them a go, that would be great.

Problem one

Compute

$L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2 x^2}dx$
$L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2 x^2}dx = \int_{a}^{\infty}\lim_{n\to\infty}\dfrac{\frac1{n} }{\frac1{n^2}+x^2}dx = \int_{a}^{\infty}\dfrac{0}{0+x^2}dx = 0$

Another Way:

$L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2 x^2}dx =\lim_{n\to\infty}(\tan^{-1} (nx)) \bigg{|}_{a}^{\infty} =$ $\lim_{n\to\infty}\left(\frac{\pi}2 -\tan^{-1} na\right) = \frac{\pi}2 - \lim_{n\to\infty}\tan^{-1} na = \frac{\pi}2 - \frac{\pi}2 = 0$

5. Originally Posted by Isomorphism
$L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2 x^2}dx = \int_{a}^{\infty}\lim_{n\to\infty}\dfrac{\frac1{n} }{\frac1{n^2}+x^2}dx = \int_{a}^{\infty}\dfrac{0}{0+x^2}dx = 0$

Another Way:

$L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2 x^2}dx =\lim_{n\to\infty}(\tan^{-1} (nx)) \bigg{|}_{a}^{\infty} =$ $\lim_{n\to\infty}\left(\frac{\pi}2 -\tan^{-1} na\right) = \frac{\pi}2 - \lim_{n\to\infty}\tan^{-1} na = \frac{\pi}2 - \frac{\pi}2{\color{red}\leftarrow\text{here}} = 0$
That is only one-thrid of the solution. Sorry, would you like me to post my solution? Or Pm it to you? I'll give you a hint, look above

6. Originally Posted by Mathstud28
Problem two

Let $a_1>a_2>a_3\cdots>a_n>0$

Furthermore let $p_1>p_2>p_3\cdots>p_n$

and $p_1+p_2+p_3+\cdots+{p_n}=1$

So then if $F(x)=\left(p_1a_1^x+p_2a_2^x+p_3a_3^x+\cdots+p_na_ n^x\right)^{\frac{1}{x}}$

Compute

$L=\lim_{x\to{0^+}}F(x)$

$L_=\lim_{x\to\infty}F(x)$

$L_2=\lim_{x\to-\infty}F(x)$
I will post my solution if the answers are right..

$L=\lim_{x\to{0^+}}F(x) = a_1^{p_1}a_2 ^{p_2} \cdots a_n^{p_n}$

$L_=\lim_{x\to\infty}F(x) = a_1$

$L_2=\lim_{x\to-\infty}F(x) = a_n$

Originally Posted by Mathstud28
That is only one-thrid of the solution. Sorry, would you like me to post my solution? Or Pm it to you? I'll give you a hint, look above
Hmm... you mean the limit is not 0?

7. Originally Posted by Isomorphism
I will post my solution if the answers are right..

$L=\lim_{x\to{0^+}}F(x) = a_1^{p_1}a_2 ^{p_2} \cdots a_n^{p_n}$

$L_=\lim_{x\to\infty}F(x) = a_1$

$L_2=\lim_{x\to-\infty}F(x) = a_n$

Hmm... you mean the limit is not 0?
Here is a huge hint, your answer should look like this

$L=\left\{\begin{array}{c}?~~\text{if}~~a?\\?~~\tex t{if}~~a?\\?~~\text{if}~~a?\end{array}\right\}$

Even bigger hint in white
arctan is odd

8. Originally Posted by Mathstud28

Here is a huge hint, your answer should look like this

$L=\left\{\begin{array}{c}?~~\text{if}~~a?\\?~~\tex t{if}~~a?\\?~~\text{if}~~a?\end{array}\right\}$

Even bigger hint in white
arctan is odd
I understand what you are trying to imply but I was wondering whats wrong with the first method

Cant I interchange the limit and the integral there?

9. Originally Posted by Isomorphism
I understand what you are trying to imply but I was wondering whats wrong with the first method

Cant I interchange the limit and the integral there?
It leaves out two answers first off, and secondly no, I do not think we can since we do not know a.

Because don't you have to have that

$\lim_{n\to\infty}\frac{n}{1+n^2x^2}\quad\forall{x} \in(a,\infty)$?

What if $0\in(a,\infty)$

Then $\lim_{n\to\infty}\frac{n}{1+n^2x^2}\ne{0}\quad\for all{x}\in(a,\infty)$

That is just one complication I see.

So I do not think an interchanging of limit and integral is permissable here.

10. Originally Posted by Mathstud28
Here are three problems I have done on other websites recently, if anyone would care to give them a go, that would be great.

Problem one

Compute

$L=\lim_{n\to\infty}\int_{a}^{\infty}\frac{n}{1+n^2 x^2}dx$

Problem two

Let $a_1>a_2>a_3\cdots>a_n>0$

Furthermore let $p_1>p_2>p_3\cdots>p_n$

and $p_1+p_2+p_3+\cdots+{p_n}=1$

So then if $F(x)=\left(p_1a_1^x+p_2a_2^x+p_3a_3^x+\cdots+p_na_ n^x\right)^{\frac{1}{x}}$

Compute

$L=\lim_{x\to{0^+}}F(x)$

$L_=\lim_{x\to\infty}F(x)$

$L_2=\lim_{x\to-\infty}F(x)$

Problem three

Prove that if $a>1$ that

$0\leq\int_a^{\infty}\frac{\sin(x)}{x}\leq\int_0^{\ infty}\frac{\sin(x)}{x}~dx$

Note

If you have seen my solution, please don't duplicate it, for that wouldn't be fun. Plus, I always want to see other points of view!

Since no one is biting I will solve the first and give hints on the others

One

$L=\lim_{n\to\infty}\int_0^{\infty}\frac{n}{1+n^2x^ 2}~dx$

So if we let

$nx=\varphi\Rightarrow\frac{d\varphi}{n}=dx$

and when $x=a\Rightarrow{\varphi=na}$

and as $x\to\infty\Rightarrow\varphi\to\infty$

So we have

$L=\lim_{n\to\infty}n\cdot\int_{na}^{\infty}\frac{\ frac{d\varphi}{n}}{1+\varphi^2}$

$=\lim_{n\to\infty}n\cdot\frac{1}{n}\int_0^{\infty} \frac{d\varphi}{1+\varphi^2}$

$=\lim_{n\to\infty}\arctan\left(\varphi\right)\bigg |_{na}^{\infty}$

$=\lim_{n\to\infty}\bigg[\frac{\pi}{2}-\arctan(na)\bigg]$

$\text{If }a>0\Rightarrow\lim_{n\to\infty}\bigg[\frac{\pi}{2}-\arctan(na)\bigg]$

$=\frac{\pi}{2}-\frac{\pi}{2}=0$

$\text{If }a=0\Rightarrow\lim_{n\to\infty}\bigg[\frac{\pi}{2}-\arctan(na)\bigg]$

$=\frac{\pi}{2}-0=\frac{\pi}{2}$

$\text{If }a<0\Rightarrow\lim_{n\to\infty}\bigg[\frac{\pi}{2}-\arctan(na)\bigg]$

$=\frac{\pi}{2}\underbrace{+}_{\text{oddness of arctan}}\frac{\pi}{2}=\pi$

$\therefore\quad\boxed{L= \left\{ \begin{array}{rcl}
0 & \mbox{if} & a>0 \\ \frac{\pi}{2} & \mbox{if} & a=0 \\ \pi & \mbox{if} & a<0

Two

$\to{0}$ Use L'hops

$\to\infty$ and $\to-\infty$ Use a sub

Three

This ones a given

11. Hello,

Similarly to the first one, compute :

$\lim_{n \to \infty} \int_a^\infty \frac{n+\sin(x)}{1+n^2x^2} ~ dx$

12. Originally Posted by Moo
Hello,

Similarly to the first one, compute :

$\lim_{n \to \infty} \int_a^\infty \frac{n+\sin(x)}{1+n^2x^2} ~ dx$
Is this going the right way? Let $a>1$. Since $\frac{n+\sin(x)}{1+(nx)^2}$ converges pointwise to $0$ and $\left|\frac{n+\sin(x)}{1+(nx)^2}\right|\leqslant\f rac{1}{1+(nx)^2}$ is integrable we can conclude by LDCT that $\lim \int_a^{\infty}\frac{n+\sin(x)}{1+(nx)^2}=0$.

Sorry if that is incorrect

13. Originally Posted by Moo
Hello,

Similarly to the first one, compute : $\lim_{n \to \infty} \int_a^\infty \frac{n+\sin(x)}{1+n^2x^2} ~ dx$
or more generally if $f(x)$ is bounded, say $|f(x)| \leq M,$ and $a>0,$ then: $\lim_{n\to\infty}\int_a^{\infty} \frac{n+f(x)}{1+n^2x^2} \ dx =0,$ because: $0 \leq \left|\int_a^{\infty} \frac{n+f(x)}{1+n^2x^2} \ dx \right| \leq \int_a^{\infty} \frac{n + M}{1+n^2x^2} \ dx=\frac{n+M}{n} \left[\frac{\pi}{2}-\tan^{-1}(na) \right].$

now apply the squeeze theorem to finish the proof. i think the case $a \leq 0$ is more intersting. you guys might be intersted in discussing it?

14. Originally Posted by NonCommAlg
or more generally if $f(x)$ is bounded, say $|f(x)| \leq M,$ and $a>0,$ then: $\lim_{n\to\infty}\int_a^{\infty} \frac{n+f(x)}{1+n^2x^2} \ dx =0,$ because: $0 \leq \left|\int_a^{\infty} \frac{n+f(x)}{1+n^2x^2} \ dx \right| \leq \int_a^{\infty} \frac{n + M}{1+n^2x^2} \ dx{\color{red}(*)}=\frac{n+M}{n} \left[\frac{\pi}{2}-\tan^{-1}(na) \right].$

now apply the squeeze theorem to finish the proof. i think the case $a \leq 0$ is more intersting. you guys might be intersted in discussing it?
I believe that we can conclude the same result. Noting that at $\color{red}(*)$ we have eveness...I think we can conjecture that based on the eveness we must just show that $\lim\int_a^{b}\frac{n+f(x)}{1+(nx)^2}dx$ converges for all finite a,b. Am I missing something or is this trivial?

15. ok, actually it's not as interesting as i thought! the answer is $\frac{\pi}{2}$ if $a=0$ and $\pi$ if $a <0.$ here's why: we have $\lim_{n\to\infty} \int_a^{\infty} \frac{n}{1+n^2x^2} \ dx=\frac{\pi}{2} - \lim_{n\to\infty} \tan^{-1}(na)=\begin{cases} 0 & \text{if} \ \ a >0 \\ \frac{\pi}{2} & \text{if} \ \ a = 0 \\ \pi & \text{if} \ \ a < 0 \end{cases}.$

on the other hand: $0 \leq \left|\int_a^{\infty} \frac{f(x)}{1+n^2x^2} \ dx \right| \leq M \int_a^{\infty} \frac{dx}{1+n^2x^2}=\frac{M}{n}\left[\frac{\pi}{2}- \tan^{-1}(na) \right].$ thus by the squeeze theorem: $\lim_{n\to\infty} \int_a^{\infty} \frac{f(x)}{1+n^2x^2} \ dx = 0, \ \forall \ a \in \mathbb{R},$ and we're done!

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