# Math Help - Small limit marathon

1. Originally Posted by Moo

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

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This one seems way too easy....

$\lim_{x \to \tfrac \pi 2} \quad 4x \tan(2x)-\frac{\pi}{\cos(2x)}=2\pi\cdot 0-\frac{\pi}{-1}=\color{red}\boxed{\pi}$

--Chris

2. @ Mathstud, #15 : it looks complicated, huh ?

Correct substitution, but what about this approximation : $\cos x \sim 1-\frac{x^2}2$ (Taylor series, that's what I think about when talking about approximations) when $x \sim 0$

$t=x-1$

--> $\lim_{t \to 0} \frac{\ln \sin(\tfrac \pi 2+\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln \cos(\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln(1-\tfrac 12 \cdot \tfrac{\pi^2 t^2}4)}{t^2} \sim -\frac{\tfrac{\pi^2t^2}{8}}{t^2}=-\frac{\pi^2}{8}$

@ Chris : Yes

3. Originally Posted by Moo
@ Mathstud, #15 : it looks complicated, huh ?

Correct substitution, but what about this approximation : $\cos x \sim 1-\frac{x^2}2$ (Taylor series, that's what I think about when talking about approximations) when $x \sim 0$

$t=x-1$

--> $\lim_{t \to 0} \frac{\ln \sin(\tfrac \pi 2+\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln \cos(\tfrac \pi 2 t)}{t^2}=\lim_{t \to 0} \frac{\ln(1-\tfrac 12 \cdot \tfrac{\pi^2 t^2}4)}{t^2} \sim -\frac{\tfrac{\pi^2t^2}{8}}{t^2}=-\frac{\pi^2}{8}$
Dangit! That probably would have been easier...oh well...same result right?

4. Originally Posted by Moo
10/
$\lim_{x \to \tfrac \pi 6} \quad \frac{(x-\tfrac \pi 6)^2}{2 \sin(x)-1}$

Let

$t=x+\frac{\pi}{6}$

So we get

$\lim_{t\to{0}}\frac{t^2}{2\sin\left(x+\frac{\pi}{6 }\right)-1}$

Now note that

$\sin\left(x+\frac{\pi}{6}\right)$

$=\sin\left(x-\frac{\pi}{3}+\frac{\pi}{2}\right)$

$=\cos\left(x-\frac{\pi}{3}\right)$

$=\frac{\cos(x)}{2}+\frac{\sqrt{3}\sin(x)}{2}$

so

$2\sin\left(x+\frac{\pi}{6}\right)=\cos(x)+\sqrt{3} \sin(x)$

So we have

$\lim_{t\to{0}}\frac{t^2}{\cos(t)+\sqrt{3}\sin(t)+1 }=0\quad\blacksquare$

5. Originally Posted by Mathstud28
Let

$t=x+\frac{\pi}{6}$

So we get

$\lim_{t\to{0}}\frac{t^2}{2\sin\left(x+\frac{\pi}{6 }\right)-1}$

Now note that

$\sin\left(x+\frac{\pi}{6}\right)$

$=\sin\left(x-\frac{\pi}{3}+\frac{\pi}{2}\right)$

$=\cos\left(x-\frac{\pi}{3}\right)$

$=\frac{\cos(x)}{2}+\frac{\sqrt{3}\sin(x)}{2}$

so

$2\sin\left(x+\frac{\pi}{6}\right)=\cos(x)+\sqrt{3} \sin(x)$

So we have

$\lim_{t\to{0}}\frac{t^2}{\cos(t)+\sqrt{3}\sin(t)+1 }=0\quad\blacksquare$
Ahahahaha...you beat me to it. I pretty much had what you had...

--Chris

6. Originally Posted by Chris L T521
Ahahahaha...you beat me to it. I pretty much had what you had...

--Chris
Quick fingers

7. Originally Posted by Moo

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

I think you are missing a criterion here. And can I pull an IMO and say the answer is (won't give it away) by the relation of the Aritmetic mean?

8. Originally Posted by Mathstud28
I think you are missing a criterion here. And can I pull an IMO and say the answer is (won't give it away) by the relation of the Aritmetic mean?
a and b > 0 ?
i don't know this relation, but i'd say no

For the previous one, there's a simpler way, but i'll wait a while

9. Originally Posted by Moo
a and b > 0 ?
$a>b\quad{a?

10. Originally Posted by Mathstud28
$a>b\quad{a?
Nothing mentioned about it, this means that your method won't work (I can see you dividing by a or b xD)

11. Originally Posted by Moo
Nothing mentioned about it, this means that your method won't work (I can see you dividing by a or b xD)
Yeah there goes that, and the Geometric Arithmetic mean thing is so nice. but here goes nothing

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

$\Rightarrow\ln\left(L\right)=\lim_{x\to{0}}\frac{\ ln\left(\frac{a^x+b^x}{2}\right)}{x}$

$=\lim_{x\to{0}}\frac{\ln\left(a^xb^x\cdot\frac{b^{-x}+a^{-x}}{2}\right)}{x}$

$=\lim_{x\to{0}}\frac{\ln(a^xb^x)}{x}+\lim_{x\to{0} }\frac{\ln\left(\frac{a^{-x}+b^{-x}}{2}\right)}{x}$

$\Rightarrow{L=\ln(ab)+\lim_{x\to{0}}\frac{\ln\left (\frac{a^{-x}+b^{-x}}{2}\right)}{x}}$

Now for the second limt

If we let $t=-x$

We get

$-\lim_{t\to{0}}\frac{\ln\left(\frac{a^t+b^t}{2}\rig ht)}{t}=-\ln(L)$

So

$\ln(L)=\ln(ab)-\ln(L)$

$\Rightarrow{2\ln(L)=\ln(ab)}$

$\Rightarrow{\ln(L)=\ln\left(\sqrt{ab}\right)}$

$\Rightarrow{L=\sqrt{ab}}\quad\blacksquare$

Hooray! That was cool!

[tex]

12. Originally Posted by Mathstud28
I think you are missing a criterion here. And can I pull an IMO and say the answer is (won't give it away) by the relation of the Aritmetic mean?
I think something is missing as well...however, I think I almost have it...but its somewhat of a pain...

--Chris

EDIT: As usual...he beats me again...

13. Originally Posted by Chris L T521
I think something is missing as well...however, I think I almost have it...but its somewhat of a pain...

--Chris

EDIT: As usual...he beats me again...

14. Hi !
Originally Posted by Moo
11/
$\lim_{x \to a} \quad \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}$
Let $f(x)=x-a\ln x$ and $g(x)=\sqrt{2ax-x^2}$

$f'(x)= 1-\frac{a}{x}$

$f''(x)=\frac{a}{x^2}$

$g'(x)=\frac{x-a}{\sqrt{2ax-x^2}}$

\begin{aligned}g''(x)&=\frac{\sqrt{2ax-x^2}-(x-a)\cdot \frac{x-a}{\sqrt{2ax-x^2}}}{2ax-x^2}\\ &=\frac{2ax-x^2-x^2+2ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\\ &=\frac{-2x^2+4ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\end{aligned}

$\frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=\frac{f(a)-f(x)}{g(a)-g(x)}\underset{a}{\sim} \frac{(x-a) f'(a)+\frac{(x-a)^2}{2}f''(a)}{(x-a) g'(a)+\frac{(x^-a)^2}{2}g''(a)}=\frac{f''(a)}{g''(a)}$ since $f'(a)=g'(a)=0$. As $f''(a)=-\frac{1}{a}$ and $g''(a)=\frac{2a^2-a^2}{a\times a^2}=\frac{1}{a}$,

$\boxed{\lim_{x \to a} \quad \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=-\frac{\frac{1}{a}}{\frac{1}{a}}=-1}$

15. Originally Posted by flyingsquirrel
Hi !

Let $f(x)=x-a\ln x$ and $g(x)=\sqrt{2ax-x^2}$

$f'(x)= 1-\frac{a}{x}$

$f''(x)=\frac{a}{x^2}$

$g'(x)=\frac{x-a}{\sqrt{2ax-x^2}}$

\begin{aligned}g''(x)&=\frac{\sqrt{2ax-x^2}-(x-a)\cdot \frac{x-a}{\sqrt{2ax-x^2}}}{2ax-x^2}\\ &=\frac{2ax-x^2-x^2+2ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\\ &=\frac{-2x^2+4ax-a^2}{\sqrt{2ax-x^2}(2ax-x^2)}\end{aligned}

$\frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=\frac{f(a)-f(x)}{g(a)-g(x)}\underset{a}{\sim} \frac{(x-a) f'(a)+\frac{(x-a)^2}{2}f''(a)}{(x-a) g'(a)+\frac{(x^-a)^2}{2}g''(a)}=\frac{f''(a)}{g''(a)}$ since $f'(a)=g'(a)=0$. As $f''(a)=-\frac{1}{a}$ and $g''(a)=\frac{2a^2-a^2}{a\times a^2}=\frac{1}{a}$,

$\boxed{\lim_{x \to a} \quad \frac{a-x-a \ln(a)+a \ln(x)}{a-(2ax-x^2)^{\tfrac 12}}=-\frac{\frac{1}{a}}{\frac{1}{a}}=-1}$
Awesome! But isn't this kind of using Cauchy's Mean Value Theorem...which is basically L'hopital's?

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