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Reply 1
kR:0×k=0limx0xcos1x=0×limx0cos1x=0\displaystyle \forall k \in \mathbb{R} : 0\times k = 0 \Rightarrow \lim_{x \to 0} x\cos \frac{1}{x} = 0 \times \lim_{x\to 0} \cos \frac{1}{x} = 0

Is this the sort of thing you were after?
Reply 2
I think it is sufficient to note that xxcos(1x)x-x \leq x\cos (\frac{1}{x}) \leq x, and to apply some sort of squeeze theorem.
Reply 3
nuodai
kR:0×k=0limx0xcos1x=0×limx0cos1x=0\displaystyle \forall k \in \mathbb{R} : 0\times k = 0 \Rightarrow \lim_{x \to 0} x\cos \frac{1}{x} = 0 \times \lim_{x\to 0} \cos \frac{1}{x} = 0

Is this the sort of thing you were after?
Hopefully not, as that ends up causing you trouble, seeing as the last limit on the RHS doesn't actually exist.

Instead, note |x cos(1/x)| <= |x| (since |cos (1/x)|<=1).
Reply 4
nuodai
kR:0×k=0limx0xcos1x=0×limx0cos1x=0\displaystyle \forall k \in \mathbb{R} : 0\times k = 0 \Rightarrow \lim_{x \to 0} x\cos \frac{1}{x} = 0 \times \lim_{x\to 0} \cos \frac{1}{x} = 0

Is this the sort of thing you were after?
That reasoning is really dodgy. For example, kR:0×k=0limx0x1x=0limx01x=0\displaystyle \forall k \in \mathbb{R} : 0\times k = 0 \Rightarrow \lim_{x \to 0} x \frac{1}{x} = 0 \cdot \lim_{x \to 0} \frac{1}{x} = 0
nuodai
kR:0×k=0limx0xcos1x=0×limx0cos1x=0\displaystyle \forall k \in \mathbb{R} : 0\times k = 0 \Rightarrow \lim_{x \to 0} x\cos \frac{1}{x} = 0 \times \lim_{x\to 0} \cos \frac{1}{x} = 0

Is this the sort of thing you were after?


I don't think that's valid, otherwise limx01=limx0x×1x=0\displaystyle \lim_{x \to 0} 1 = \lim_{x \to 0} x \times \dfrac{1}{x} = 0, obviously wrong.
Use Sandwich Rule with y=x and y= -x.
Reply 7
By the way, this is a Cambridge University Tripos Exam question from first year Engineering.

We have never done sandwich rule, so there must be another way.
ricman3
By the way, this is a Cambridge University Tripos Exam question from first year Engineering.

We have never done sandwich rule, so there must be another way.


- == it is probably the most intuitive / 'simple' techniques you could learn when dealing with limits.
Reply 9
Surely cos(1/x) will tend towards 1, so xcos(1/x) will diverge to the same infinity...?
I don't see why any calculation is required, it's simple to see straight away.
the cosine part takes a value between 0 and 1 always, it is pretty irrelevant what it is, as you're multiplying it by a number very close to 0. A very small fraction multiplied by a number between 1 and 0 is clearly tending to 0.
Elementric
I don't see why any calculation is required, it's simple to see straight away.
the cosine part takes a value between 0 and 1 always, it is pretty irrelevant what it is, as you're multiplying it by a number very close to 0. A very small fraction multiplied by a number between 1 and 0 is clearly tending to 0.


That's the idea but makes a weak proof on it's own. Also, x is not defined at 0, but the function has a limit as it tends to 0.
silent ninja
That's the idea but makes a weak proof on it's own. Also, x is not defined at 0, but the function has a limit as it tends to 0.


It just seems like more of a 1 mark 'write down the answer' question to me.
Elementric
It just seems like more of a 1 mark 'write down the answer' question to me.
Seeing as the OP said he knew the answer was 0, and wanted an algebraic justification, simply writing down the answer is not really adequate.
Reply 14
I think it was a 3 mark question. I agree it is kind of obvious. Anyway, the squeeze theorem is probably fine, thanks for the help.
Reply 15
Don't you need to do the cos power expansion series??? (cos x = 1 - x^2/2! + x^4/4! + ... etc...) Sub in 1/x instead of x and see what happens...
Just a thought, that's how I'd go about it
DFranklin
Seeing as the OP said he knew the answer was 0, and wanted an algebraic justification, simply writing down the answer is not really adequate.


Hence my brief explanation in a previous post. Thanks for reading.
Reply 17
you get x - 1/(2!x) + 1/(4!x^3) +...
not sure what you can do from there.
mr_t_42
Don't you need to do the cos power expansion series??? (cos x = 1 - x^2/2! + x^4/4! + ... etc...) Sub in 1/x instead of x and see what happens...
Just a thought, that's how I'd go about it


:eek3:
Reply 19
ricman3
you get x - 1/(2!x) + 1/(4!x^3) +...
not sure what you can do from there.


Hmm thats what I got.

No, I'm afraid i don't have any ideas about what to do next... :-(

Whenever ive done something like this before, its just expanded into something that's quite nice to work out but im stumped now, lol