TT's Maths Thread

[TrueTears]

Is there even any solutions, TT?


My reasoning, may be epic fail ><

If

Then





When ...............[1],







As , substitute x = 0.

..........[2]



Substitute into [1]:







Substituting a = 1, b = 2 into [2]:

Equation is undefined.

Doing the same for

We find that:

...but and are the only solutions.

Substituting them into when , we find that the equation is again, undefined.

Hence no solutions.

I am 90% sure my reasoning is flawed...

Don't think you can substitute

Needs some other substitution.

Since the division law doesn't apply here, we need to somehow find a substitution that gets rid of the on the bottom...

[Damo17]

Find and such that

Need to introduce something new here, but what lol

Thanks

Rationalising the Numerator gives:


Thus this is true only if the numerator tends to 0, so .

So we sub in and get:



thus and .

[TrueTears]

Since the division law doesn't apply here...

It doesn't? Can you elaborate on this, I'm interested.

Ok, goes back to hopeless pondering..

Because you have x on the denominator.

Find and such that

Need to introduce something new here, but what lol

Thanks

Rationalising the Numerator gives:


Thus this is true only if the numerator tends to 0, so .

So we sub in and get:



thus and .

But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use

[/0]

this might work...



, as





Let's say...







(taking a cue from one of kamil's earlier solutions )

[Damo17]

But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use

http://en.wikipedia.org/wiki/Division_by_zero

When you have the form where f(x) and g(x) approach 0 as x approaches 0 then this limit may equal a real or infinite value, or may not exist at all.

EDIT:
I should also add that I used L'Hopital's rule, thus the limit did not reach .

[TrueTears]

But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use

http://en.wikipedia.org/wiki/Division_by_zero

When you have the form where f(x) and g(x) approach 0 as x approaches 0 then this limit may equal a real or infinite value, or may not exist at all.

EDIT:
I should also add that I used L'Hopital's rule, thus the limit did not reach .

Cool, but division by 0 in this is still not well-defined?

Also I try not to use L'hopital's theorem whenever possible It is too cheap ^.^ That's why I tried to go for more wishful methods haha

[Damo17]

But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use

http://en.wikipedia.org/wiki/Division_by_zero

When you have the form where f(x) and g(x) approach 0 as x approaches 0 then this limit may equal a real or infinite value, or may not exist at all.

EDIT:
I should also add that I used L'Hopital's rule, thus the limit did not reach .

Cool, but division by 0 in this is still not well-defined?

Also I try not to use L'hopital's theorem whenever possible It is too cheap ^.^ That's why I tried to go for more wishful methods haha

It is not well-defined.

I do understand your hatred of L'Hopital's rule (I also dislike it) but unfortunately that is the only way to do this question as far as I can see.

[TrueTears]

But if numerator approaches 0 and you limit x to 0 for the denominator we again have a undefined case.

So you can not use

http://en.wikipedia.org/wiki/Division_by_zero

When you have the form where f(x) and g(x) approach 0 as x approaches 0 then this limit may equal a real or infinite value, or may not exist at all.

EDIT:
I should also add that I used L'Hopital's rule, thus the limit did not reach .

Cool, but division by 0 in this is still not well-defined?

Also I try not to use L'hopital's theorem whenever possible It is too cheap ^.^ That's why I tried to go for more wishful methods haha

It is not well-defined.

I do understand your hatred of L'Hopital's rule (I also dislike it) but unfortunately that is the only way to do this question as far as I can see.

Ah okay, awesome thanks, it was a good method nonetheless, but yeah I was looking for more of a substitution method like /0

[/0]

[TrueTears]

.

[/0]

d'oh, now everyone knows

[TrueTears]

d'oh, now everyone knows

THA.....

[/0]

d'oh, now everyone knows

THA.....

FKN.....

[TrueTears]

Hmm

[/0]

Perhaps think about it a bit longer first, but here's an idea

=\frac{4}{6}lim_{x \to 0} \frac{\sin 4x}{4x} \times \frac{6x}{\sin 6x}