How would you prove this limit theorem? (2 Viewers)

Driftonian · Jan 9, 2024

Any help would be appreciated. Many thanks!

WeiWeiMan · Jan 9, 2024

Driftonian said:
Any help would be appreciated. Many thanks!

le hospital or sub in small values of x (neither of which are in syllabus i think)

Bored of studying · Jan 9, 2024

WeiWeiMan said:
le hospital or sub in small values of x (neither of which are in syllabus i think)

No u fucking retard ass bitch
u take - out from numerator, then u use double angle formula, take the 2 out and bobs your fucking uncle you dissappointment bitch

Average Boreduser · Jan 9, 2024

Bored of studying said:
No u fucking retard ass bitch
u take - out from numerator, then u use double angle formula, take the 2 out and bobs your fucking uncle you dissappointment bitch

yeah I think this mans approach is pre good. you see the limit then turns to -2limsinx/x*limsinx and taking limx->0sinx=0 thererfore the rhs=0.

Driftonian · Jan 9, 2024

Average Boreduser said:
yeah I think this mans approach is pre good. you see the limit then turns to -2limsinx/x*limsinx and taking limx->0sinx=0 thererfore the rhs=0.

Why are you able to split the multiplication of the limits?

Bored of studying · Jan 9, 2024

Driftonian said:
Why are you able to split the multiplication of the limits?

its a property of limits hun lim(f(x)*g(x))=limf(x)*limg(x)

Average Boreduser · Jan 9, 2024

Driftonian said:
Why are you able to split the multiplication of the limits?

Bored of studying said:
its a property of limits hun lim(f(x)*g(x))=limf(x)*limg(x)

^

WeiWeiMan · Jan 9, 2024

Driftonian said:
Why are you able to split the multiplication of the limits?

It's not in syllabus don't listen to Bored Of Studying.

Average Boreduser · Jan 9, 2024

WeiWeiMan said:
It's not in syllabus don't listen to Bored Of Studying.

nah i think it is. its just noting limsinx/x=1 and limsinx=0. I think its basically what u said, but I think you'd lose marks if you don't indicate limfxgx=limfxlimgx

WeiWeiMan · Jan 9, 2024

Average Boreduser said:
nah i think it is. its just noting limsinx/x=1 and limsinx=0. I think its basically what u said, but I think you'd lose marks if you don't indicate limfxgx=limfxlimgx

yeah prove that tho cuz i don't think it's explicitly stated in syllabus

Average Boreduser · Jan 9, 2024

WeiWeiMan said:
yeah prove that tho cuz i don't think it's explicitly stated in syllabus

Yeah I mean I think it just uses intuition. Like its not explicitly on there, but applying laws of limits, you can see as x->0; sinx->0 and thats ur ans. Kinda is indirectly related/

cossine · Jan 10, 2024

Driftonian said:
Why are you able to split the multiplication of the limits?

While other post have given answer I thought would add more detail. There things called indeterminant forms. Basically if you cannot evaluate a limit using substitution then you have indeterminant form. (E.g., 0/0, 0^0, infinity/infinity, 1^infinity, infinity - infinity, 0*infinity)

So in the question if we substitute theta = 0 in the question we will get 0/0 which is undefined.

In the case we have an indeterminant form then we can't apply the limit laws.

A simple example: lim x->infinity x - x. Applying the limit laws we would get infinity -infinity which is undefined. However by simplifying the expression we can see the limit is clearly 0. A minor note, don't confuse indeterminant forms with arithmetic. If you want ask a limit with indeterminant form infinity - infinity equals 0. But you shouldn't ask why does infinity -infinity equal 0.

Unfortunately, the proof of the limit laws going by memory involved epsilon delta definition of limits. For high school mathematics you are only expected to know the intuitive definition not the rigorous epsilon delta definition.

https://math.libretexts.org/Bookshelves/Calculus/Calculus_(OpenStax)/02:_Limits/2.03:_The_Limit_Laws

Drongoski · Jan 10, 2024

$\lim _{\theta \to 0} \frac {cos\theta - 1}{\theta} = \lim_{\theta \to 0} (-\frac {1-cos\theta}{\theta} )= -\lim _{\theta \to 0}\frac {2sin^2(\frac {{\theta}}{2})}{\theta}\\ \\ =-\lim_{\theta \to 0} (sin \frac {\theta}{2}) \times \lim _{\theta \to 0} (\frac {sin \frac {\theta}{2}}{\frac {\theta}{2}})\\ \\ = - \lim _{\frac {\theta}{2} \to 0} sin \frac {\theta}{2} \times \lim _{\frac {\theta}{2} \to 0}\frac {sin \frac {\theta}{2}}{\frac {\theta}{2}}\\ \\ = - 0 \times 1 = 0$

Note:
Since most of you would not have learnt the theory of limits, you may not fully appreciate my steps.

Jedhcurtis76 · Jan 10, 2024

Drongoski said:
$\lim _{\theta \to 0} \frac {cos\theta - 1}{\theta} = \lim_{\theta \to 0} (-\frac {1-cos\theta}{\theta} )= -\lim _{\theta \to 0}\frac {2sin^2(\frac {{\theta}}{2})}{\theta}\\ \\ =-\lim_{\theta \to 0} (sin \frac {\theta}{2}) \times \lim _{\theta \to 0} (\frac {sin \frac {\theta}{2}}{\frac {\theta}{2}})\\ \\ = - \lim _{\frac {\theta}{2} \to 0} sin \frac {\theta}{2} \times \lim _{\frac {\theta}{2} \to 0}\frac {sin \frac {\theta}{2}}{\frac {\theta}{2}}\\ \\ = - 0 \times 1 = 0$

Note:
Since most of you would not have learnt the theory of limits, you may not fully appreciate my steps.

Beautifully done!

For anyone confused over why

Screen Shot 2024-01-10 at 10.13.48 am.png

Please find the proof (using Squeeze Theorem) attached!

Drongoski · Jan 10, 2024

Jedhcurtis76 said:
Beautifully done!
For anyone confused over why

View attachment 42052

Please find the proof (using Squeeze Theorem) attached!

In Ext 1 Maths, you can assume:

$\lim _{\theta \to 0} \frac {sin \theta}{\theta} = 1$

Jedhcurtis76 · Jan 10, 2024

Drongoski said:
In Ext 1 Maths, you can assume:

$\lim _{\theta \to 0} \frac {sin \theta}{\theta} = 1$

Yeh I know, but it’s cool to know WHY. Then you can really believe it from there.

ZakaryJayNicholls · Jan 10, 2024

Driftonian said:
Any help would be appreciated. Many thanks!

Really easy way is to split the fraction (limit of product is product of limits):

lim x->0 (cosx-1/x) = lim x->0 (1/x) * lim x->0 (cosx-1) = inf * 0 = 0

liamkk112 · Jan 10, 2024

ZakaryJayNicholls said:
Really easy way is to split the fraction (limit of product is product of limits):

lim x->0 (cosx-1/x) = lim x->0 (1/x) * lim x->0 (cosx-1) =0 * 0 = 0

but lim x-> 0 of 1/x is infinity tho

liamkk112 · Jan 10, 2024

liamkk112 said:
but lim x-> 0 of 1/x is infinity tho

not to mention it doesnt exist since 1/x goes to -inf from 0- and inf from 0+

liamkk112 · Jan 10, 2024

ZakaryJayNicholls said:
inf times zero = zero

in general this is not true

How would you prove this limit theorem? (2 Viewers)

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