Limits question (2 Viewers)

[Mumma]

LIM (cos[x] - 1) / x
x->0

How would you do this?

 

[robbo_145]

let f(x) = cos[x] - 1
f'(x) = sin[x]
f(0) = 0

let g(x) = x
g'(x) = 1
g(0) = 0

thus applying l'hopital
LIM (cos[x] - 1) / x
x->0
= lim f(x)/g(x)
x->0
= lim f'(x)/g'(x)
x->0

gives
lim sin[x]/1 = 0
x->0

can't remember if you get taught L'hopital in 4 unit, but its handy to know

 

[zeek]

wouldn't you do something like this...

Lim (cos[x]-1)/x -------> Lim(cos[x]/x - 1/x)

as x-> 0, cos[x]-> 1, 1/x -> inf

.: Lim (cos[x]-1)/x ------->Lim(inf. - inf.)
------->0

im not sure if that makes any sense though

 

[STx]

wouldn't you do something like this...

Lim (cos[x]-1)/x -------> Lim(cos[x]/x - 1/x)

as x-> 0, cos[x]-> 1, 1/x -> inf

.: Lim (cos[x]-1)/x ------->Lim(inf. - inf.)
------->0

im not sure if that makes any sense though

^ that is correct according to maple

 

[vafa]

Done

 

[Mumma]

wouldn't you do something like this...

Lim (cos[x]-1)/x -------> Lim(cos[x]/x - 1/x)

as x-> 0, cos[x]-> 1, 1/x -> inf

.: Lim (cos[x]-1)/x ------->Lim(inf. - inf.)
------->0

im not sure if that makes any sense though

Yes, I did think of 'infinity - infinity' but I dont know if that is valid.

let f(x) = cos[x] - 1
f'(x) = sin[x]
f(0) = 0

let g(x) = x
g'(x) = 1
g(0) = 0

thus applying l'hopital
LIM (cos[x] - 1) / x
x->0
= lim f(x)/g(x)
x->0
= lim f'(x)/g'(x)
x->0

gives
lim sin[x]/1 = 0
x->0

can't remember if you get taught L'hopital in 4 unit, but its handy to know

Ill look into it, thank you.

Motherfucker what the fuck.
lim g(x)/f(x) = lim g'(x)/f'(x)
holy shit. That looks to be very useful, thank you.


One problem
I was using the lim x->0 (cos[x]-1)/x to PROVE d(sinx)/dx = cos x
[ take LIM as lim h-> 0]

d(sin x).dx = LIM (sin[x+h] - sinx)/h

= LIM (sinx cosh + cosx sinh - sin x)/h

= LIM sinx(cosh-1)/h + cosx LIM sinh/h {----> LIM sinh/h = 1}

= sinx LIM (cosh - 1)/h + cosx

I KNOW that LIM (cosh - 1)/h = 0, I just have to prove it!
I also know there is an alternate way of proving d(sinx)/dx by using sums to products and whatever, but I want to do it this way.

 

[airie]

^ that is correct according to maple

I somehow don't think that, when x and y are two functions of which the values approach infinity at the limit, lim(x-y)=0...I mean, infinity is not a number o.0

My try is to first apply the half-angle rule, so (cos x)-1 = -2sin²(x/2). Therefore,
lim [(cos x)-1]/x
x->0
= lim [-2sin²(x/2)]/x
x->0
= lim -2sin(x/2)*[sin(x/2)]/x
x->0
= lim -2sin(x/2)*(1/2)*[sin(x/2)]/(x/2)
x->0
= lim -sin(x/2) * lim [sin(x/2)]/(x/2)
x->0 x->0
=0 * 1
=0.

EDIT: Yeah, used the fact that lim [sin(x/2)]/(x/2) = 1 as x approaches 0, same as vafa. Wow, that took me so long to type up

 

[Mumma]

Hey thanks vafa for that, Im going to read it now.

EDIT: Argh so many replies
Anyway Vafa you say 2(sinx)² = 1-cosx?? That doenst look right. Still reading all these replies...

 

[vafa]

Sorry about that
but you need to use the same idea and i gurantee that it is not going to make any difference. This method is explained more by Gaha in his 3 unit 50 HSc tips.

 

[Mumma]

Wow I never actually knew that
LIM g(x)f(x) = LIM g(x) * LIM f(X)
I actually did reach that sin²[x/2] thing, I just didnt know that law applied! Thanks so much!

By the way, when we reach
-2 LIM sin[x/2] /x * LIM sin[x/2]
since we know LIM sin[x/2] = 0, cant we say the whole thing is 0 without using sinx/x = 1?

 

[vafa]

Since you are talking about Extension1, then the answer is yes.

 

[robbo_145]

My try is to first apply the half-angle rule, so (cos x)-1 = -2sin²(x/2).

ahh i knew there would be a 4 unit way to do it, nice solution

lim g(x)/f(x) = lim g'(x)/f'(x)

the above condition only holds if g(x) and f(x) either both tend to infinity; or both tend to 0 as in the above case

 

[SeDaTeD]

Depends on your definition of cos(x) .

Using the power series definition, cos(x) = 1 - x^2/2! + x^4/4! - x^6/6! + ...
cos(x) - 1 = -x^2/2! + x^4/4! - x^6/6! + ...
(cos(x)-1)/x = -x + x^3/4! - x^5/6! + ...

Therefore, as x-> 0, (cos(x)-1)/x -> 0.

However, I doubt they'd accept that in the hsc.

Also note, for l'Hopital's rule, you would have to prove it before you use it, as it is not taught in the course. It can only be applied when the numerator and denominator approaches something of the form 0/0 or inf/inf.

Proof:
If f(x) and g(x) are functions and f'(a) and g'(a) exist, with f(x), g(x) -> 0 as x -> a, then
lim x->a f(x)/g(x)
= lim x->a [f(x) - f(a)]/[g(x) - (a)], as f(a) = g(a) = 0 (*)
= lim x-> a {[f(x) - f(a)]/(x-a)} / {[g(x) - g(a)]/(x-a)}
= {lim x-> a [f(x) - f(a)]/(x-a)}/{lim x->a [g(x) - g(a)]/(x-a)}
= f'(a)/g'(a), if the derivatives exist

(*) The fact that the derivatives exist at x=a implies that the functions are continuous at x=a. Hence f(a) = lim x->a f(x) = 0 and likewise for g.

You can treat the inf/inf case by letting h = 1/x and taking the limit as h-> 0.

Edit: I've noticed another thing. Try to find the derivative of cos(x) using first principles. You will see that it involves using the fact that (cosx -1)/x goes to 0 as x -> 0. So using l'Hopital's rule in this case may constitute circular reasoning (though, if you're lucky, your teacher may not notice). Best way to go is to use airie's method.

 

[Mumma]

Edit: I've noticed another thing. Try to find the derivative of cos(x) using first principles. You will see that it involves using the fact that (cosx -1)/x goes to 0 as x -> 0. So using l'Hopital's rule in this case may constitute circular reasoning (though, if you're lucky, your teacher may not notice). Best way to go is to use airie's method.

Yes, that was the problem I was initially trying to solve, and I realised the circular reasoning and noted that out. I dont think Ill end up using l'Hopital's rule, but it is interesting regardless.

 

[airie]

By the way, when we reach
-2 LIM sin[x/2] /x * LIM sin[x/2]
since we know LIM sin[x/2] = 0, cant we say the whole thing is 0 without using sinx/x = 1?

I was gonna do that at first, but you have x as the denominator for one of the terms so I suppose you can't really just sub x=0 in

 

[Raginsheep]

You're not subbing in x = 0, Your finding the behaviour of the curve as it approaches 0 which is slightly different.

at x = 0, your function could equal any random value and it would have no effect on your limit.

 

[airie]

You're not subbing in x = 0, Your finding the behaviour of the curve as it approaches 0 which is slightly different.

at x = 0, your function could equal any random value and it would have no effect on your limit.

Hmm. You're right.

 

[SeDaTeD]

Actually, come to think of it, the fact that a derivative exists implies the function is continous anyway (check, for those who know the formal definition of a limit). We would then only require the condition that the derivatives exist for f and g at x=a. I should add that.

 

[Raginsheep]

Yes but to be honest, no HSC student is going into go to such detail nor do they need to.

 

[SeDaTeD]

Agreed, but it's there for those who want it.