How would you prove this limit theorem? (1 Viewer)

[ZakaryJayNicholls]

Note1: The limits do exist, they simply exist uniquely from each side.
Note2: The concern as to the answer being zero is "does 1/x go to infinity faster than cosx-1 goes to zero", 1/0.1=10, 1/0.01=100 eg linear, cos0.1-1 =-0.0049, cos0.01-1=-0.0000499 eg faster than linear.
This means the cos term goes to zero faster than the 1/x term goes to infinity, thus the whole thing tends to zero.

 

[liamkk112]

Note1: The limits do exist, they simply exist uniquely from each side.
Note2: The concern as to the answer being zero is "does 1/x go to infinity faster than cosx-1 goes to zero", 1/0.1=10, 1/0.01=100 eg linear, cos0.1-1 =-0.0049, cos0.01-1=-0.0000499 eg faster than linear.
This means the cos term goes to zero faster than the 1/x term goes to infinity, thus the whole thing tends to zero.

note1: yes they do definetly exist, however the notation lim x->0 is ambigouous as to which direction x is approaching from, meaning that in this case it would most likely not be a good idea to simply write that the limit goes to infinity at least without clarification, even in hsc where this might be overlooked. a good idea might be to at least clarify the situation and take the limits in both directions, showing that the negative sign doesnt matter in this case as we are converging to zero.

note2: this is true that cosx-1 dominates 1/x in the limit as x goes to zero, however this was not acknowledged in the answer properly. simply saying inf x 0 = 0 is most definetly incorrect, my teachers would cry or laugh if they saw that i wrote that.

 

[ZakaryJayNicholls]

note1: yes they do definetly exist, however the notation lim x->0 is ambigouous as to which direction x is approaching from, meaning that in this case it would most likely not be a good idea to simply write that the limit goes to infinity at least without clarification, even in hsc where this might be overlooked. a good idea might be to at least clarify the situation and take the limits in both directions, showing that the negative sign doesnt matter in this case as we are converging to zero.

note2: this is true that cosx-1 dominates 1/x in the limit as x goes to zero, however this was not acknowledged in the answer properly. simply saying inf x 0 = 0 is most definetly incorrect, my teachers would cry or laugh if they saw that i wrote that.

I did say that I was presenting a really easy way, a quick and novel method, which works in this context.

Complaining about the fact that it's not rigorous enough for your specific taste is a waste of time.

 

[liamkk112]

I did say that I was presenting a really easy way, a quick and novel method, which works in this context.

Complaining about the fact that it's not rigorous enough for your specific taste is a waste of time.

im not? there is a difference between a quick and novel method and something that literally makes no sense. inf x 0 is literally one of the indeterminate forms, every calculus student should know that, no matter if they are in advanced, extension 2 or university maths. that isnt particularly rigourous, that is basically the fundamental difference between a limit expression making sense or not. a quick and novel method would be the one that was presented here before, using double angle trig to simplify the expression. that method is three lines, i would say that is quite fast, while also making sense.

your method does technically work and is also quite succinct, however only with a couple of extra clarifications that would make the method much clearer as i pointed out, as currently it is ambiguous. i was not criticising your solution pointlessly or complaining about a lack of rigour that matches my "taste", a quick aside that clarifies the domination in the limit and checks both sides of the limit is integral to the working out to the solution. my response added on to these notes that should have likely been in the original solution, as if a student posted your original solution it would be dubious without your notes and possibly the additional clarification that i made to note 1, that is nothing to do with rigour, but to do with indeterminate forms, something that is fundamental to all students who are learning about limits, again regardless of skill level (this is why i said that my teachers would cry or laugh if they saw that i wrote that, i would likely be made fun of as this is one of the first things you learn about calculus; this was for the most part a joke, i am sorry if it came off as something else).

 

[ZakaryJayNicholls]

im not? there is a difference between a quick and novel method and something that literally makes no sense. inf x 0 is literally one of the indeterminate forms, every calculus student should know that, no matter if they are in advanced, extension 2 or university maths. that isnt particularly rigourous, that is basically the fundamental difference between a limit expression making sense or not. a quick and novel method would be the one that was presented here before, using double angle trig to simplify the expression. that method is three lines, i would say that is quite fast, while also making sense.

your method does technically work and is also quite succinct, however only with a couple of extra clarifications that would make the method much clearer as i pointed out, as currently it is ambiguous. i was not criticising your solution pointlessly or complaining about a lack of rigour that matches my "taste", a quick aside that clarifies the domination in the limit and checks both sides of the limit is integral to the working out to the solution. my response added on to these notes that should have likely been in the original solution, as if a student posted your original solution it would be dubious without your notes and possibly the additional clarification that i made to note 1, that is nothing to do with rigour, but to do with indeterminate forms, something that is fundamental to all students who are learning about limits, again regardless of skill level (this is why i said that my teachers would cry or laugh if they saw that i wrote that, i would likely be made fun of as this is one of the first things you learn about calculus; this was for the most part a joke, i am sorry if it came off as something else).

Your method does technically work and is also quite succinct - correct. Now lets just end this here.

 

[liamkk112]

Your method does technically work and is also quite succinct - correct. Now lets just end this here.

bro i legit said that u need the notes and clarifications i made for it to make sense and without them the original answer doesnt make sense... but alright.

 

[Average Boreduser]

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

Bro when u x anythign by 0 its =0 wym
infinite isnt a definite number. for all we know it could be 100 or 10000. By multiplying it by 0, its always gonna be 0.

Edit : this resp. sounds agressive asf mbmb ^^^^ I don't mean it to be rude lol. just to clarify. (I could be wrong i guess)

 

[liamkk112]

Bro when u x anythign by 0 its =0 wym
infinite isnt a definite number. for all we know it could be 100 or 10000. By multiplying it by 0, its always gonna be 0.

Edit : this answers sounds agressive asf mbmb ^^^^ I don't mean it to be rude lol. just to clarify. (I could be wrong i guess)

its a common misconception but it is an indeterminate form, this is exactly why i clarified it lmao... i wasnt trying to complain or critcise the other guy i just wanted everyone to understand it well . it dont sound agressive it just sounds like a question

think about it this way: your argument could be reversed. infinity times any number is also infinity, by the same idea that 0 x anything is also zero. so i can also say that infinity x 0 = infinity. we have conflicting answers, so it must be indeterminate.

naturally there are functional examples where this is the case. f(x) = (e^x)(e^-x) is one, in the limit that x -> inf. notice in this case that e^x will go to infinity, e^-x goes to 0, so in the limit that x-> inf f(x) = inf x 0. but upon simplifying f(x) = 1, so naturally in the limit f(x) goes to 1. again there are conflicting answers, and this can only occur if the limit is indeterminate; so this means that infinity x 0 must be an indeterminate form. im sure there are many other examples/proofs of this, but this is one that i could find.

 

[Average Boreduser]

its a common misconception but it is an indeterminate form, this is exactly why i clarified it lmao... i wasnt trying to complain or critcise the other guy i just wanted everyone to understand it well . it dont sound agressive it just sounds like a question

think about it this way: your argument could be reversed. infinity times any number is also infinity, by the same idea that 0 x anything is also zero. so i can also say that infinity x 0 = infinity. we have conflicting answers, so it must be indeterminate.

naturally there are functional examples where this is the case. f(x) = (e^x)(e^-x) is one, in the limit that x -> inf. notice in this case that e^x will go to infinity, e^-x goes to 0, so in the limit that x-> inf f(x) = inf x 0. but upon simplifying f(x) = 1, so naturally in the limit f(x) goes to 1. again there are conflicting answers, and this can only occur if the limit is indeterminate; so this means that infinity x 0 must be an indeterminate form. im sure there are many other examples/proofs of this, but this is one that i could find.

oh. ok

 

[Luukas.2]

I advise strongly against making unsupported assertions about indeterminate forms.

Unjustified statement (dangerous in marking terms):

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Reasonable working:

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Unjustified statement (dangerous in marking terms):

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Reasonable working:

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or:

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