Small limit marathon

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• Jul 20th 2008, 10:22 AM
Moo
Small limit marathon
...not as long as Mathstud's. So feel free to create a new one (Tongueout)
Some (if not all) are easy, so @ the masters (not http://www.mathhelpforum.com/math-he...s/masters.html), pardon me for insulting your abilities and please lend normal people 10 more minutes than you need for solving them (Rofl)

Of course, $\color{red}\huge \text{NO L'HOSPITAL'S RULE}$ :D
(you may be able to use the gradient definition). No power series is needed but approximations could be needed in some limits.

I think solutions can all be less than 5 or 6 lines...

1/
$\lim_{x \to \tfrac \pi 2} \quad \left(\frac \pi x -1\right)^{\tan(x)}$

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2/
$\lim_{x \to \tfrac \pi 2} \quad 4x \tan(2x)-\frac{\pi}{\cos(2x)}$

Solution #1 (Chris L T521) & book's answer

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3/
$\lim_{x \to 0} \quad \frac{(x+27)^{\tfrac 13}-3}{(x+16)^{\tfrac 14}-2}$

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4/
$\lim_{x \to 0} \quad \left(\frac{a^x+b^x}{2}\right)^{\tfrac 1x}$

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5/
$\lim_{x \to 0} \quad \frac{e^x-e^{\sin(x)}}{x-\sin(x)}$

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6/
$\lim_{x \to 0} \quad \frac{x^{\sqrt{x}}}{(\sqrt{x})^x}$

Solution #1 (Mathstud)

Alternative

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7/
$\lim_{x \to 1} \quad \frac{2}{1-x^2}-\frac{3}{1-x^3}$

Solution #1 (Mathstud)

Alternative

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8/
$\lim_{x \to 1} \quad \frac{\ln(\sin(\tfrac \pi 2 x))}{(x-1)^2}$

Solution #1 (Mathstud)

Alternative

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9/
$\lim_{x \to 1} \quad \frac{1-x^\alpha}{\ln(x)}$

Solution #1 (Mathstud)

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10/
$\lim_{x \to \tfrac \pi 6} \quad \frac{(x-\tfrac \pi 6)^2}{2 \sin(x)-1}$

Solution #1 (Mathstud)

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11/
$\lim_{x \to a} \quad \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}$

Have fun (Nod)
• Jul 20th 2008, 10:39 AM
Mathstud28
Quote:

Originally Posted by Moo
...not as long as Mathstud's. So feel free to create a new one (Tongueout)
Some (if not all) are easy, so @ the masters (not http://www.mathhelpforum.com/math-he...s/masters.html), pardon me for insulting your abilities and please lend normal people 10 more minutes than you need for solving them (Rofl)

Of course, $\color{red}\huge \text{NO L'HOSPITAL'S RULE}$ :D
(you may be able to use the gradient definition). No power series is needed but approximations could be needed in some limits.

I think solutions can all be less than 5 or 6 lines...

Have fun (Nod)

Wait! I can't use power series?!? Are you serious? What about the Root and Ratio test trick?

Until you answer here is 9

$\lim_{x\to{1}}\frac{1-x^{\alpha}}{\ln(x)}$

$=-\lim_{x\to{1}}\int_0^{\alpha}x^{y}~dy$

$=-\int_0^{\alpha}\lim_{x\to{1}}x^y~dy$

$=-\int_0^{\alpha}~dy$

$=-\alpha$
• Jul 20th 2008, 10:42 AM
Moo
No Root & Ratio test trick either ^^ If you want to, it's ok, but I don't really like this method (never saw it in books (Surprised))
No integral is needed either (Rofl) but yes you can do it this way.

There are various ways of solving the limits. But as far as I can see, no use of big tricks like integrals or power series.

(hey, don't quote the whole post !!!! it's painful for internet connections :D)
• Jul 20th 2008, 10:46 AM
Mathstud28
Quote:

Originally Posted by Moo
No Root & Ratio test trick either ^^ If you want to, it's ok, but I don't really like this method (never saw it in books (Surprised))

I guess I am an an inovator (Smirk):p

Quote:

No integral is needed either (Rofl) but yes you can do it this way.

There are various ways of solving the limits.

(hey, don't quote the whole post !!!! it's painful for internet connections :D)

Then you must have some kind of method you prefer! I would love more restrictions since I can do these mostly with one or two methods. But if you say you can only use _____ method it will be harder! Your call though. This is entirely your thread.

And I deleted some of the quote (Nod)

--------------------------------------------------------------------------------------------------

By the way, is that the correct answer?
• Jul 20th 2008, 10:50 AM
Moo
Quote:

Originally Posted by Mathstud28
I guess I am an an inovator (Smirk):p

Looks weird though :p

Quote:

Then you must have some kind of method you prefer! I would love more restrictions since I can do these mostly with one or two methods. But if you say you can only use _____ method it will be harder! Your call though. This is entirely your thread.
There's no particular method I advocate, I read it in a book and I liked them all.
But I think using integrals and power series sometimes just kills too fast the problem :D

Quote:

And I deleted some of the quote (Nod)
Thanks :D

Quote:

By the way, is that the correct answer?
I forgot to tell ya, yes it is ^^

I'll put a red sign rather when it is false than when it is correct (Wink)
• Jul 20th 2008, 10:52 AM
Mathstud28
Quote:

Originally Posted by Moo
Looks weird though :p

There's no particular method I advocate, I read it in a book and I liked them all.
But I think using integrals and power series sometimes just kills too fast the problem :D

Use substitutions, approximations and TRICKS and your INTUITION (Rofl)

Thanks :D

I forgot to tell ya, yes it is ^^

I'll put a red sign rather when it is false than when it is correct (Wink)

Ok great! I am going to get started on these...by the way you have a typo in problem 8 I think.
• Jul 20th 2008, 11:11 AM
Mathstud28
Quote:

Originally Posted by Moo
6/
$\lim_{x \to 0} \quad \frac{x^{\sqrt{x}}}{(\sqrt{x})^x}$

$\lim_{x\to{0}}\frac{x^{\sqrt{x}}}{\sqrt{x}^x}$

Let $x=u^2$

So as

$x\to{0}\Rightarrow{u\to{0}}$

So we have

$\lim_{u\to{0}}\frac{u^{2u}}{u^{u^2}}$

Now let us consider these two limits seperately

Firstly

$L=\lim_{u\to{0}}u^{2u}$

$\Rightarrow\frac{1}{2}\ln\left(L\right)=\lim_{u\to {0}}u\ln(u)$

Now let $\frac{1}{t}=u$

So we have that s $u\to{0}\Rightarrow{t\to\infty}$

Giving us

$\frac{1}{2}\ln\left(L\right)=-\lim_{t\to\infty}\frac{\ln(t)}{t}$

Now I do not know if this suffices but $\ln(t)\prec{t}$

$\Rightarrow\lim_{t\to\infty}\frac{-\ln(t)}{t}=0$

So

$\frac{1}{2}\ln\left(L\right)=0\Rightarrow{L=1}$

Now consider

$L_1=\lim_{u\to{0}}u^{u^2}$

So

$\ln\left(L_1\right)=\lim_{u\to{0}}u^2\ln(u)$

Once again let $u=\frac{1}{t}$

Giving us

$\ln\left(L_1\right)=\lim_{t\to\infty}\frac{-\ln(t)}{t^2}$

Now if you don't allow precs then

$\exists{N}\backepsilon\forall{t>N}\quad{0\leq{\ln( t)}{\sqrt{t}}}$

This implies that

$\exists{N}\backepsilon\forall{t>N}\quad{0\leq\frac {\ln(t)}{t^2}\leq\frac{1}{t^{\frac{3}{2}}}}$

This now implies that

$0\leq\lim_{t\to\infty}\frac{\ln(t)}{t^2}\leq\lim_{ t\to\infty}\frac{1}{t^{\frac{3}{2}}}$

$\Rightarrow\quad{0\leq\lim_{t\to\infty}\frac{\ln(t )}{t^2}\leq{0}}$

This implies by the squeeze Theorem that

$\lim_{t\to\infty}\frac{\ln(t)}{t^2}=0$

$\therefore\quad\ln\left(L_1\right)=0$

$\Rightarrow\quad{L_1=1}$

So now seeing that

$\lim_{u\to{0}}\frac{u^{u^2}}{u^{2u}}$

$=\frac{\lim_{u\to{0}}u^{u^2}}{\lim_{u\to{0}}u^{2u} }$

$=\frac{L}{L_1}$

$=\frac{1}{1}$

$=1$

So

$\boxed{\lim_{x\to{0}}\frac{x^{\sqrt{x}}}{\sqrt{x}^ x}=1}\quad\blacksquare$
• Jul 20th 2008, 11:20 AM
Moo
Your problem for 6/ is that you assume the limit exists (L and L_1).

But isn't 0^0 undefined ?

I have a different answer (Wink)
• Jul 20th 2008, 11:22 AM
Mathstud28
Quote:

Originally Posted by Moo
7/
$\lim_{x \to 1} \quad \frac{2}{1-x^2}-\frac{3}{1-x^3}$

$\lim_{x\to{1}}\frac{2}{1-x^2}-\frac{3}{1-x^3}$

To this one, all we must do is combine fractions

$\frac{2}{1-x^2}-\frac{3}{1-x^3}$

$=\frac{2}{(1-x)(1+x)}-\frac{3}{(1-x)(1+x+x^2)}$

$=\frac{2(1+x+x^2)}{(1-x)(1+x)(1+x+x^2)}-\frac{3(1+x)}{(1-x)(1+x)(1+x+x^2)}$

$=\frac{2x^2-x+1}{(1-x)(1+x)(1+x+x^2)}$

$=\frac{-(2x+1)}{(1+x)(1+x+x^2)}$

So

$\lim_{x\to{0}}\frac{2}{1-x^2}-\frac{3}{1-x^3}$

$=-\lim_{x\to{0}}\frac{2x+1}{(1+x)(1+x+x^2)}$

$=\frac{-1}{2}\quad\blacksquare$
• Jul 20th 2008, 11:23 AM
Mathstud28
Quote:

Originally Posted by Moo
Your problem for 6/ is that you assume the limit exists (L and L_1).

But isn't 0^0 undefined ?

I have a different answer (Wink)

Are you saying the answer isn't one?
• Jul 20th 2008, 11:26 AM
Moo
Quote:

Originally Posted by Mathstud28
Are you saying the answer isn't one?

What is the meaning of different ? (Rofl)
Yes, the answer isn't 1 ^^ (at least that's what the book says and it seems correct :D)

Don't wander too far.

------------------
It's ok for 7/
• Jul 20th 2008, 11:30 AM
Mathstud28
Quote:

Originally Posted by Moo
What is the meaning of different ? (Rofl)
Yes, the answer isn't 1 ^^ (at least that's what the book says and it seems correct :D)

Don't wander too far.

------------------
It's ok for 7/

Well after you contested my answer I checked it on maple and my TI-89

If the limit was in fact

$\lim_{x\to{0}}\frac{x^{\sqrt{x}}}{\sqrt{x}^x}$ then the answer is one.
• Jul 20th 2008, 11:32 AM
Moo
Quote:

Originally Posted by Mathstud28
Well after you contested my answer I checked it on maple and my TI-89

If the limit was in fact

$\lim_{x\to{0}}\frac{x^{\sqrt{x}}}{\sqrt{x}^x}$ then the answer is one.

(Yawn)
I'm sincerely sorry, I misread that. It was the ln which was 0 (Rofl)

$\ln(y)=x^{1/2} \ln(x)-\frac x2 \ln x$, which tends to 0 when x tends to 0 (Tongueout)

Thus lim y=1
• Jul 20th 2008, 11:34 AM
Mathstud28
Quote:

Originally Posted by Moo
(Yawn)
I'm sincerely sorry, I misread that. It was the ln which was 0 (Rofl)

$\ln(y)=x^{1/2} \ln(x)-\frac x2 \ln x$, which tends to 0 when x tends to 0 (Tongueout)

Thus lim y=1

Don't worry about it...come here and I'll tell you a secret...I make lots of clarical errors while doing math problems....shh! Keep that one under your cap. (Wink)
• Jul 20th 2008, 12:27 PM
Mathstud28
Quote:

Originally Posted by Moo
8/
$\lim_{x \to 1} \quad \frac{\ln(\sin(\tfrac \pi 2 x))}{(x-1)^2}$

First we note that

$\sin\left(\frac{\pi{x}}{2}\right)=\sqrt{1-\cos^2\left(\frac{\pi{x}}{2}\right)}$

So

$\ln\left(\sin\left(\frac{\pi{x}}{2}\right)\right)= \frac{1}{2}\ln\left(1-\cos^2\left(\frac{\pi{x}}{2}\right)\right)$

So we have

$\frac{1}{2}\lim_{x\to{1}}\frac{\ln\left(1-\cos^2\left(\frac{\pi{x}}{2}\right)\right)}{(x-1)^2}$

Now let

$t=x-1$

So as $x\to{1}\Rightarrow{t\to{0}}$

So we have

$\lim_{t\to{0}}\frac{\ln\left(1-\cos^2\left(\frac{\pi(t+1)}{2}\right)\right)}{t^2}$

Now we know that as x goes to zero

$\ln(1-x^2)\sim{-x^2}$

So now here comes the trick

$\cos^2\left(\frac{\pi(t+1)}{2}\right)$

$=\bigg[\cos\left(\frac{\pi{t}}{2}+\frac{\pi}{2}\right)\bi gg]^2$

$=\bigg[-\sin\left(\frac{\pi{t}}{2}\right)\bigg]^2$

$=\bigg[\sin\left(\frac{\pi{t}}{2}\right)\bigg]^2$

Now we know that

$\sin\left(\frac{\pi{t}}{2}\right)\sim\frac{\pi{t}} {2}$

So

$\sin^2\left(\frac{\pi{t}}{2}\right)\sim\frac{\pi^2 t^2}{4}$

So we have that

$\frac{1}{2}\lim_{t\to{0}}\frac{\ln\left(1-\cos^2\left(\frac{\pi(t+1)}{2}\right)\right)}{t^2} \sim\frac{1}{2}\lim_{t\to{0}}\frac{-\frac{\pi^2t^2}{4}}{t^2}=\frac{-\pi^2}{8}$

$\therefore\quad\boxed{\lim_{x\to{1}}\frac{\ln\left (\sin\left(\frac{\pi{x}}{2}\right)\right)}{(x-1)^2}=\frac{-\pi^2}{8}}\quad\blacksquare$
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