Author Topic: TT's Maths Thread (Read 146011 times) Tweet Share

kamil9876 · « **Reply #660 on:** January 09, 2010, 06:09:58 pm »

u could do an $\epsilon-\delta$ proof

The idea behind that is that if we compute values $\frac{-1}{x^2}$ for x-values below -0.5, we can make $|\frac{-1}{x^2}-4|$ less than any desired positive value. Ie we can minimize the 'approximation error' as much as we like.

TrueTears · « **Reply #661 on:** January 09, 2010, 06:22:37 pm »

lol kk how about

$\lim_{x \to 0^+} \left( \frac{1}{x} - \frac{1}{|x|}\right)$

If $x \ge 0 \implies |x| = x$

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

But I sketched the graph of $\frac{1}{x} - \frac{1}{|x|}$ , there is no graph on the right side of 0...

nvm I got it, it was cause the graph was y = 0 lolz

TrueTears · « **Reply #662 on:** January 09, 2010, 09:12:35 pm »

$\lim_{x \to 0} \lfloor \cos(x) \rfloor \ \forall \ -\pi \le x \le \pi$

Book says it's undefined.

Shouldn't it be 0?

Since: $\lim_{x \to 0^-} \lfloor \cos(x) \rfloor =\lim_{x \to 0^+} \lfloor \cos(x) \rfloor = 0$

TrueTears · « **Reply #663 on:** January 09, 2010, 09:31:41 pm »

Quote from: Damo17 on January 09, 2010, 09:31:24 pm

Quote from: TrueTears on January 09, 2010, 09:12:35 pm
$\lim_{x \to 0} \lfloor \cos(x) \rfloor \ \forall \ -\pi \le x \le \pi$

Book says it's undefined.

Shouldn't it be 0?

Since: $\lim_{x \to 0^-} \lfloor \cos(x) \rfloor =\lim_{x \to 0^+} \lfloor \cos(x) \rfloor = 0$

Does the question have absolute signs?

Nah just the floor function.

TrueTears · « **Reply #664 on:** January 09, 2010, 09:37:28 pm »

Quote from: kamil9876 on January 09, 2010, 09:36:06 pm

book is right only limit from the positive side is 0. Limit from the negative side is -1 since for all x such that $-\pi\leq x< 0$ the $-1\leq cosx<0$ so -1 is the greatest integer less than it. (rememebr, the value at x=0 is irrelevant to the limit).

No it is still 0 from the left.

kamil9876 · « **Reply #665 on:** January 09, 2010, 09:44:12 pm »

ahh eyah shit i was thinking of sinx since it would be correct for that one

. Yeah you are correct they're wrong.

TrueTears · « **Reply #666 on:** January 09, 2010, 09:45:57 pm »

Quote from: kamil9876 on January 09, 2010, 09:44:12 pm

ahh eyah shit i was thinking of sinx since it would be correct for that one . Yeah you are correct they're wrong.

Thanks kamilz

TrueTears · « **Reply #667 on:** January 09, 2010, 10:11:20 pm »

$\lim_{x \to 2}\left(\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\right)$

I've tried rationalising the numerator and denominator... none works, can't use L'hopital's theorem... what to do?

brightsky · « **Reply #668 on:** January 09, 2010, 10:22:36 pm »

You can use L'Hospital's rule. I pretty sure.

Let $f(x) = \sqrt{6-x} - 2$ and $g(x) = \sqrt{3-x}-1$ .

Using L'Hospital's rule,

$\frac {f'(x)}{g'(x)} = \sqrt{\frac{x-3}{x-6}}$

$\implies \lim_{x\to2} \sqrt{\frac{x-3}{x-6}}$

Substitute $x = 2$ ,

$\implies \sqrt{\frac{2-3}{2-6}} \implies \sqrt{\frac{-1}{-4}} \implies \sqrt {\frac{1}{4}} \implies \frac {1}{2}$

TrueTears · « **Reply #669 on:** January 09, 2010, 10:24:53 pm »

Quote from: brightsky on January 09, 2010, 10:22:36 pm

You can use L'Hospital's rule. I pretty sure.

Yeah I know you can use it, but I specifically avoid it whenever I can because it's not fun and simply ruins the question.

I'm after a more elegant solution.

brightsky · « **Reply #670 on:** January 09, 2010, 10:31:53 pm »

Oh ok, my bad..hmmm...

TrueTears · « **Reply #671 on:** January 09, 2010, 11:56:20 pm »

Quote from: TrueTears on January 09, 2010, 10:11:20 pm

$\lim_{x \to 2}\left(\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\right)$

Actually I think I got it.

Let $x = u+2$

$\sqrt{6-x} = \sqrt{4-u}$

$\left(4-u\right)^{\frac{1}{2}} = \sqrt{4}\left(1-\frac{u}{4}\right)^{\frac{1}{2}} = \sqrt{4} \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}(1)^{\frac{1}{2}-k}\left(\frac{-u}{4}\right)^k = 2 \left[ \binom{\frac{1}{2}}{0}\left(\frac{-u}{4}\right)^0 + \binom{\frac{1}{2}}{1}\left(\frac{-u}{4}\right)^1 + \binom{\frac{1}{2}}{2}\left(\frac{-u}{4}\right)^2 + \cdots \right] = 2\left[ 1 - \frac{u}{8} - \frac{u^2}{128} + \cdots \right] = 2 - \frac{u}{4} - \frac{u^2}{64} + \cdots$

$\sqrt{3-x} = \sqrt{1-u}$

$\left(1-u\right)^{\frac{1}{2}} = \sum_{k=0}^{\infty}\binom{\frac{1}{2}}{k}(1)^{r-k}(-u)^k = \binom{\frac{1}{2}}{0}(-u)^0 + \binom{\frac{1}{2}}{1}(-u)^1 + \binom{\frac{1}{2}}{2}(-u)^2 + \cdots = 1 - \frac{u}{2} - \frac{u^2}{8} + \cdots$

$\implies \lim_{x \to 2}\left(\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}\right) = \lim_{u \to 0}\left(\frac{2 - \frac{u}{4} - \frac{u^2}{64} + \cdots-2}{1 - \frac{u}{2} - \frac{u^2}{8} + \cdots -1}\right) =\lim_{u \to 0}\left(\frac{\left(- \frac{u}{4} - \frac{u^2}{64} + \cdots\right)\left(\frac{1}{u}\right)}{\left(- \frac{u}{2} - \frac{u^2}{8} + \cdots\right)\left(\frac{1}{u}\right)}\right) = \lim_{u \to 0} \left(\frac{-\frac{1}{4} - \frac{u}{64} + \cdots}{-\frac{1}{2} - \frac{u}{8} + \cdots}\right) = \frac{1}{2}$

kamil9876 · « **Reply #672 on:** January 09, 2010, 11:59:30 pm »

Ironically, binomial theorem is more sophisticated/advanced/harder to prove than L'hopital's rule.

TrueTears · « **Reply #673 on:** January 10, 2010, 12:00:23 am »

fuck l'hopital's rule, that cheap shit

/0 · « **Reply #674 on:** January 10, 2010, 12:00:50 am »

LOL
I see calculus isn't liked very much around here

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