Question

prove that lim x->0 cos(1/x) does not exist.

Answer 1

\[\lim_{x\rightarrow 0}\cos(\frac{1}{x})\]
we can start by looking at what 1/x is. If that is undefined then we can assume cos(undefined) = undefined. (or DNE)
\[lim_{x \rightarrow 0^+}\frac{1}{x}=\infty \]
and
\[lim_{x\rightarrow0^-}\frac{1}{x}=-\infty\]
therefore
\[\lim_{x\rightarrow0}\frac{1}{x}=undefined\]
and
\[\lim_{x\rightarrow0}\cos(undefined)=undefined\]

Answer 2

no right and left hand limits yet. can't use them.

Answer 3

how about this: letx_n = 1/2pi(n), lim x_n = 0
but limf(x_n) = lim cos(2pin) = 1 since lim x_n does not equal lim f (x_n) thr limit does not exist

Answer 4

the limit does not exist

Answer 5

I think there is a way to show
\[\cos\frac{1}{0} = \sin\infty\]
and lim x->infty sin(x) = undefined.

Answer 6

no l'hopitals rule yet either,

Answer 7

Well, that is just annoying.