/sci/ - Science & Math

>>5686623
You aren't just "subbing in 0 for h", firstly.

I think the idea is that those limits have been determined earlier.

>>5686626
My understanding was you solve algebraically so that you can then sub in your limit without causing errors.

>>5686623
If you solve the two last limits by using L'Hopitals theorem you get 1 and 0, respectively.

>>5686623
>>5686623
each of those has been calculated before, look up their calculations on google to see how.

>>5686638
>>5686639
Ah ok. Thanks for the info.

>>5686638
Don't listen to this guy, that's circular logic.

Somehow, without using L'Hôpital (because of circular proof), they got that <span class="math">\lim_{h \to 0}\frac{cos(h)-1}{h}=0[/spoiler] and <span class="math">\lim_{h \to 0}\frac{sin(h)}{h}=1[/spoiler]. Plot <span class="math">\frac{sin(x)}{x}[/spoiler] and <span class="math">\frac{cos(x)-1}{x}[/spoiler] in GeoGebra and assume that their limits are 1 and 0 respectively

>>5686623
this question has bothered me my entire uni life how the shit do you do those limits without circular logic

>>5686840
If you define the sine and cosine functions as power series, those limits follow pretty easily.

>>5686623
>>5686812
>>5686840
You can solve both those limits using geometry in the unit circle, or at least make them plausible that way.
If you don't like geometry you can as >>5686846
said define sine and cosine their taylor series and prove it that way. Although then you need to prove that the powers series correspond to the geometric interpretation.
Lastly, you can prove the sinx/x limit using the fact that sinx < x < tanx when 0 < x < pi/2 and solve it using the squeeze theorem! (cosx-1)/x can probably be proven similarly.

>>5686846
thats not how you define them bro

you define them as the solutions to the differential equations f''(x) + f(x) = 0 with the conditions f(0)=1, f'(0)=0 and f(0)=0, f'(0)=1

>>5686840
it's not rigorous at all, but "for small values of $x$", you can approximate $\sin{x}=x,\cos{x}={1-x}$

>>5686638
>wants to find the derivative of cos(x)
>uses, that the derivative of cos(x) is -sin(x) in his proof
genius

>>5686623

<div class="math">\frac{d^n}{dx^n}cos(x)= cos(x+\frac{n\pi}{2}) n\epsilon \mathbb{R}</div>

>>5686846
I dont want to define them that way though, because then I have to prove that the unir circle has co-ords (cos(x),sin(x)) from the series definition and that sounds difficult. But this thread has made me look up the geometric proof now so I'm happy.
So now: define sine/cos in terms of the circle -> prove those limits geometrically -> prove derivatives -> prove taylor series. All nice and consistent and lovely

>>5686866
<div class="math">n\epsilon \mathbb{Z}</div>

https://www.khanacademy.org/math/calculus/limits_topic/squeeze_theorem/v/proof--lim--sin-x--x

He goes nice and slow for you.

For the cosine part, just multiply and divide by (cos(h) + 1)

>>5686881
<div class="math">\frac{cos(h)-1}{h}\cdot \frac{cos(h)+1}{cos(h)+1}</div>

<div class="math">=\frac{cos^2(h)-1}{h(cos(h)+1)} = \frac{-sin(h)}{h(cos(h)+1)}</div>

The bottom will still approach zero.

Just look at proof wiki and be done with it:
http://www.proofwiki.org/wiki/Derivative_of_Cosine_Function

>>5686884
No.

You have sin^2x on th etop.

make that sinh/h times sinh/divided by the cosine mess.

The video proves that sinh/h = 1 , so you have 1 * 0 = 0. You just use the sinh/h limit twice is all.

>>5686888
Sorry, left out the squared..

<div class="math">\frac{-sin^2(h)}{h(cos(h)+1)} = \frac{-sin^2(h)}{h}\cdot \frac{1}{(cos(h)+1)}</div>

Is that what you are suggesting?
It is wrong.
Because,
<div class="math">\lim_{h\rightarrow 0}\frac{-sin^2(h)}{h}\neq 1</div>

<div class="math">\lim_{h\rightarrow 0}\frac{-sin^2(h)}{h}=0</div>

this is an intuitive non-rigorous look but it will hopefully help you understand more

That's calc 1 stuff man. Thosr are special trig limits. Sinx/x is equal to 1. Cosx -1/x is equal to zero. I think these are proved using squeeze theorum I'm not sure though