WTF!!!!! Limit (1 Viewer)

[blackbunny]

lim as x--->0 (1-cos(x)) / x^2


if cos was squared i would have no problems , but its not WTF!!! how the F do u do this

 

[Slidey]

1=cos^2(x/2)+sin^2(x/2)
cosx=cos^2(x/2)-sin^2(x/2)
1-cosx=2sin^2(x/2)
2sin^2(x/2)= TOP
and
x^2=4(x/2)^2= BOTTOM

So your question is:
lim [2sin^2(x/2)]/[4(x/2)^2]
x->0
=
1/2(lim sin^2(x/2)/(x/2)^2)
x--->0
=1/2 * 1 * 1=1/2

 

[blackbunny]

in 2nd line of working wer the hell does cosx and - come from. 3rd line of woking WTF!!. 6th line of working wer this 4(x/2)^2 comes from. Tnx

 

[Slidey]

Second line:
cos(a+b)=cosacosb-sinasinb
When b=a,
cos2a=cosacosa-sinasina=cos^2(a)-sin^2(a)

Now, x=x/2 + x/2, so:
cosx=cos(x/2 + x/2)=cos^2(x/2)-sin^2(x/2)

Third line:
1-cosx=1-(cos^2(x/2)-sin^2(x/2))=1-cos^2(x/2) + sin^2(x/2)
But from the fundamental identity, 1-cos^2(x/2)=sin^2(x/2), so:
1-cos^2(x/2) + sin^2(x/2) = sin^2(x/2) + sin^2(x/2) = 2sin^2(x/2)

Sixth(?) line:
x^2=4(x/2)^2

 

[blackbunny]

till not sureaboutthe 3rd line 1-cos^2(x/2) + sin^2(x/2) = sin^2(x/2) + sin^2(x/2) = 2sin^2(x/2) ........!!!!!!!!!!!!!!!!!

 

[Slidey]

OK, well what's 1-cos2x?

 

[blackbunny]

1/2(lim sin^2(x/2)/(x/2)^2)
x--->0
=1/2 * 1 * 1=1/2


how did u get .5 *1 *1 what did u put in for x????

 

[Slidey]

It's the fundamental limit:

lim sinx/x
x-->0
= 1

Now

lim sin^2(x)/x^2
x-->0

Is simply two limits multiplied together like so:
( lim sinx/x ) * ( lim sinx/x )
Both as x-->0
Which is 1 * 1

1/2 comes from 2/4.

I think you should take a look at your limit laws:

g(x), h(x) are functions
c is a constant

Lim A*g(x)
x->c
=
A * Lim g(x)
x->c

Lim h(x)*g(x)
x->c
=
[Lim g(x)] * [Lim h(x)]
Both as x->c

Postscript: This would be so much easier with l'Hospital's rule.

 

[Trev]

We had a similar question to this in the exam we did on thursday!
Similar to a question in maths in focus textbook, trigonometric function topic (forget which exercise)

 

[Slidey]

By L'Hospital's rule:
Numerator with limit applied = 0, denominator with limit applied = 0, so:
take derivative of top on derivative of bottom and apply the limit:
lim sinx/2x
x->0
=1/2

That's like 2 lines of working!

 

[JamiL]

slide is that on the sylubus... n if not can u use it

 

[shafqat]

l'hopitals is very useful
but its not on the syllabus
so unfortunately it cant be used

 

[Slidey]

I've heard otherwise, shafqat. What's your reasoning?

 

[shafqat]

mainly because someone stumbled across it in a class test
and dr pender told us it cant be used because its not in the syllabus
i myself have no idea

 

[FinalFantasy]

l'hopitals is very useful
but its not on the syllabus
so unfortunately it cant be used

but i made a thread a while ago, and ppl were saying u can use anything in maths even if it's not in the syllabus?

 

[shafqat]

come on
in this case u really cant use l'hopitals cos it makes the question too trivial
it would disadvantage people who werent taught this method
you have to set limits somewhere (no pun intended), and thats what the syllabus does

 

[FinalFantasy]

but shouldn't ppl who went to the trouble of going to learn it get some credit for it
"it would disadvantage people who werent taught this method"
slide rule probably wasn't "taught" it though, it doesn't get "taught" anyway.
so not disadvantage, just a credit to ppl who are keen and went to learn more!

but anyway... so is it true u really can't use things outside da syllabus?
last time ppl said u can and stuff -_-
now im confused

 

[shafqat]

well all kudos to those who learn outside the syllabus
but consider this: some of the circle geometry questions are just corollaries or related to well known results, which many olympiad students know
can they just right: "proven, because of blah"?

 

[FinalFantasy]

hmm..
so now...
in ext. 1 and ext. 2 u can only use things within the syllabus right?

 

[shafqat]

pretty much