Small limit marathon

**wingless** · July 20th 2008, 12:34 PM

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.

In order the limit to exist, $\lim_{x\to{0^+}}\frac{x^{\sqrt{x}}}{\sqrt{x}^x}=\l im_{x\to{0^-}}\frac{x^{\sqrt{x}}}{\sqrt{x}^x}$ should be true. But $\lim_{x\to{0^-}}\frac{x^{\sqrt{x}}}{\sqrt{x}^x}$ doesn't exist because of that square root. Some books assume that limits should be evaluated as one-sided limits at the endpoints. In my opinion, the question should've been $\lim_{x\to{0^+}}\frac{x^{\sqrt{x}}}{\sqrt{x}^x}$ to avoid confusion.

**Mathstud28** · July 20th 2008, 12:34 PM

Originally Posted by Moo

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

$e^x\sim{1+x}$

$e^{\sin(x)}\sim{1+\sin(x)}$

$\therefore\quad{e^x-e^{\sin(x)}\sim{x-\sin(x)}}$

$\therefore\quad\lim_{x\to{0}}\frac{e^x-e^{\sin(x)}}{x-\sin(x)}\sim\lim_{x\to{0}}\frac{x-\sin(x)}{x-\sin(x)}=1\quad\blacksquare$

**Moo** · July 20th 2008, 12:37 PM

Originally Posted by Mathstud28

Awesome! But isn't this kind of using Cauchy's Mean Value Theorem...which is basically L'hopital's?

Isn't it sort of a Taylor series expansion ? I wasn't very clear with "approximations" stuff... But once again feel free to use the method you like (I just point you towards one in particular lol). The only rule is no l'hospital's ^^

@ flyingsquirrel : great

Here is an alternative method :

Let $t=x-a$

The limit is now $\lim_{t \to 0} \quad \frac{-t+a \ln(1+\tfrac ta)}{a-\sqrt{a^2-t^2}}$

$-t+a \ln(1+\tfrac ta) \sim -t+a(\tfrac ta-\tfrac{t^2}{2a^2})=-\frac{t^2}{2a}$

$\frac1{a-\sqrt{a^2-t^2}}=\frac{t^2}{a+\sqrt{a^2-t^2}} \sim \frac{t^2}{2a}$

Thus the limit is...

@ Mathstud, #32 : good ^^

**Mathstud28** · July 20th 2008, 12:39 PM

Originally Posted by Moo

Isn't it sort of a Taylor series expansion ?

No

**flyingsquirrel** · July 20th 2008, 12:52 PM

Originally Posted by Mathstud28

Awesome! But isn't this kind of using Cauchy's Mean Value Theorem...which is basically L'hopital's?

Yes it's quite close from l'Hospital's rule. (but I didn't use it

)

Originally Posted by Mathstud28

Isn't it sort of a Taylor series expansion ?

No

Yes it is, I didn't use Cauchy's mean value theorem.

Originally Posted by Moo

Here is an alternative method :

Let $t=x-a$

The limit is now $\lim_{t \to 0} \quad \frac{-t+a \ln(1+\tfrac ta)}{a-\sqrt{a^2-t^2}}$

$-t+a \ln(1+\tfrac ta) \sim -t+a(\tfrac ta-\tfrac{t^2}{2a^2})=-\frac{t^2}{2a}$

$\frac1{a-\sqrt{a^2-t^2}}=\frac{t^2}{a+\sqrt{a^2-t^2}} \sim \frac{t^2}{2a}$

Thus the limit is...

... more rapidly found using this method.

**Mathstud28** · July 20th 2008, 01:09 PM

Originally Posted by Moo

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

Let $\frac{1}{\tan(x)}=t\Rightarrow{\arctan\left(\frac{ 1}{t}\right)=x}$

So as $x\to\frac{\pi}{2}\Rightarrow{t\to{0}}$

So we have

$\lim_{t\to{0}}\left(\frac{\pi}{\arctan\left(\frac{ 1}{t}\right)}-1\right)^{\frac{1}{t}}$

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

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

Now suppose the limit exists and is equal to L then

$\ln\left(L\right)=\lim_{t\to{0}}\frac{\ln\left(\fr ac{-2t-\pi}{2t-\pi}\right)}{t}$

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

$\sim\lim_{t\to{0}}\frac{\frac{-4t}{2t-\pi}}{t}$

$=\frac{4}{\pi}$

$\Rightarrow{L=e^{\frac{4}{\pi}}}\quad\blacksquare$

I have a feeling that this is much easier...I hate trig functions and non-zero limits xD

EDIT :I ommited a step that might cause confusion

$\arctan\left(\frac{1}{t}\right)=\frac{\pi}{2}-\arctan(t)$

and

$\arctan(t)\sim{t}$

**Moo** · July 20th 2008, 01:17 PM

Originally Posted by flyingsquirrel

... more rapidly found using this method.

@ Mathstud : weird substitution, but it works ^^
especially thinking about this acrtan approximation

Try substitution $t=x-\tfrac \pi 2$ and then you'll get, after taking the logarithm and after some steps (2 or 3), to $-\frac{1}{\tan(t)} \left(\ln(1-\tfrac{2t}{\pi})-\ln(1+\tfrac{2t}{\pi})\right)$

$\tan(t) \sim t$

etc ^^

**Mathstud28** · July 20th 2008, 02:07 PM

Originally Posted by Moo

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

Well I was trying to give someone else a chance, but I am getting tired of waiting and I want to finish this off

$\lim_{x\to{0}}\frac{(x+27)^{\frac{1}{3}}-3}{(x+16)^{\frac{1}{4}}-2}$

$=\lim_{x\to{0}}\frac{(x+27)^{\frac{1}{3}}-3}{x}\cdot\lim_{x\to{0}}\frac{x}{(x+16)^{\frac{1}{ 4}}-2}$

$=L\cdot{L_1}$

Now by definition of the derivative

$L=\lim_{x\to{0}}\frac{(x+27)^{\frac{1}{3}}-3}{x}=\left((x+27)^{\frac{1}{3}}-3\right)'\bigg|_{x=0}=32$

and

$\frac{1}{L_1}=\lim_{x\to{0}}\frac{(x+16)^{\frac{1} {4}}-2}{x}=\left((x+16)^{\frac{1}{4}}-2\right)'\bigg|_{x=0}=27$

So

$L\cdot{L_1}=\frac{32}{27}$

$\therefore\quad\boxed{\lim_{x\to{0}}\frac{(x+27)^{ \frac{1}{3}}-3}{(x+16)^{\frac{1}{4}}-2}=\frac{32}{27}}$

And we end this marathon.

EDIT: Wow, I sounded really really arrogant in the beginning. Sorry everyone

unintentional I assure you.

**galactus** · July 20th 2008, 02:31 PM

I like seeing others do these, but I thought I would throw my hat in.

I will take a stab at one. But I like to watch mathstud, Moo, and others go at them. Then I can absorb some of their 'useful tricks'.

Here is a method I used for this particular integral some years back. I have seen it before.

It just entails a little algebra and e.

Here is #4 using a little algebra and a substitution.

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

Let $t=\frac{1}{x}, \;\ x=\frac{1}{t}$

Then, we note that as ${x\to 0}$ , $t\to {\infty}$

Make the sub and we get:

$\lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}+b^{\frac{1}{t }}}{2}\right)^{t}$

Now, a little algebra and 'e'; add 1 and subtract 1:

$\lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}+b^{\frac{1}{t }}}{2}+1-1\right)$

$\lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}+b^{\frac{1}{t }}-2}{2}+1\right)^{t}$

Because of asymptotics that +1 is just a 'hanger on' and we can rewrite as such using e:

$\lim_{t\to {\infty}}e^{t\left(\frac{a^{\frac{1}{t}}+b^{\frac{ 1}{t}}-2}{2}\right)}$

$\lim_{t\to {\infty}}e^{\frac{1}{2}\left(\frac{a^{\frac{1}{t}} +b^{\frac{1}{t}}-2}{\frac{1}{t}}\right)}$

$\lim_{n\to {\infty}}e^{\frac{1}{2}\left(\frac{a^{\frac{1}{t}}-1}{\frac{1}{t}}+\frac{b^{\frac{1}{t}}-1}{\frac{1}{t}}\right)}$

Now, the limit, which is the exponent of e, can be taken:

$\frac{1}{2}\lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}-1}{\frac{1}{t}}+\frac{b^{\frac{1}{t}}-1}{\frac{1}{t}}\right)$

Now, the limits here I won't bother deriving. I will use it as a 'lemma'
$\lim_{t\to {\infty}}(\frac{a^{\frac{1}{t}}-1}{\frac{1}{t}})=\lim_{t\to {\infty}}(a^{\frac{1}{t}}-1)t=ln(a)$

Analogous with b, which is ln(b).

So, we end up with:

$e^{\frac{1}{2}(ln(a)+ln(b))}$

$e^{ln(\sqrt{ab})}$

Which leads to $\sqrt{ab}$

**Mathstud28** · July 20th 2008, 02:56 PM

Originally Posted by galactus

I like seeing others do these, but I thought I would throw my hat in.

I will take a stab at one. But I like to watch mathstud, Moo, and others go at them. Then I can absorb some of their 'useful tricks'.

Here is a method I used for this particular integral some years back. I have seen it before.

It just entails a little algebra and e.

Here is #4 using a little algebra and a substitution.

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

Let $t=\frac{1}{x}, \;\ x=\frac{1}{t}$

Then, we note that as ${x\to 0}$ , $t\to {\infty}$

Make the sub and we get:

$\lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}+b^{\frac{1}{t }}}{2}\right)^{t}$

Now, a little algebra and 'e'; add 1 and subtract 1:

$\lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}+b^{\frac{1}{t }}}{2}+1-1\right)$

$\lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}+b^{\frac{1}{t }}-2}{2}+1\right)^{t}$

Because of asymptotics that +1 is just a 'hanger on' and we can rewrite as such using e:

$\lim_{t\to {\infty}}e^{t\left(\frac{a^{\frac{1}{t}}+b^{\frac{ 1}{t}}-2}{2}\right)}$

$\lim_{t\to {\infty}}e^{\frac{1}{2}\left(\frac{a^{\frac{1}{t}} +b^{\frac{1}{t}}-2}{\frac{1}{t}}\right)}$

$\lim_{n\to {\infty}}e^{\frac{1}{2}\left(\frac{a^{\frac{1}{t}}-1}{\frac{1}{t}}+\frac{b^{\frac{1}{t}}-1}{\frac{1}{t}}\right)}$

Now, the limit, which is the exponent of e, can be taken:

$\frac{1}{2}\lim_{t\to {\infty}}\left(\frac{a^{\frac{1}{t}}-1}{\frac{1}{t}}+\frac{b^{\frac{1}{t}}-1}{\frac{1}{t}}\right)$

Now, the limits here I won't bother deriving. I will use it as a 'lemma'
$\lim_{t\to {\infty}}(\frac{a^{\frac{1}{t}}-1}{\frac{1}{t}})=\lim_{t\to {\infty}}(a^{\frac{1}{t}}-1)t=ln(a)$

Analogous with b, which is ln(b).

So, we end up with:

$e^{\frac{1}{2}(ln(a)+ln(b))}$

$e^{ln(\sqrt{ab})}$

Which leads to $\sqrt{ab}$

Awesome method Galactus! I think its really cool how you always find ways to relate things to e

Just for those who are wondering abo ut Galactus's Lemma

$\lim_{t\to\infty}\left(a^{\frac{1}{t}}-1\right)t$

Let $t=\frac{1}{x}$

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

$=\lim_{x\to{0}}\frac{1+\ln(a)x+\cdots-1}{x}=\ln(a)$

**Marine** · July 20th 2008, 09:40 PM

thank you, it was pleasure for me to see so many beautiful and various solutions to difficult limits

now I really got how powerful de l'Hospital's rule actually is

, saving lots of work

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