Find the limit of (cos1 / x) / X when x tends to 0

Find the limit of (cos1 / x) / X when x tends to 0

Limit[Cos[1/x]/x,x→0],
If t = 1 / x
Limit[Cos[1/x]/x,x→0]
=Limit[t Cos[t],t→∞],
Let t = 2n π,
Limit[t Cos[t],t→∞]
=Limit[t Cos[2nπ],n→∞]
=Limit[t,n→∞]
= ∞
Let t = (2n + 1 / 2) π,
Limit[t Cos[t],t→∞]
=Limit[t Cos[(2n+1/2)π],n→∞]
= 0,
therefore
Limit[Cos[1/x],x→0],
The limit does not exist

When x tends to infinity, what is the limit of limx (cos1 / x-1)?

The original formula = LIM (x tends to infinity) (cos1 / x-1) / (1 / x). Using lobida's law, LIM (x tends to infinity) - sin (1 / x) = 0

Why does LIM (when x approaches 0) (2 * x * sin1 / X - CON1 / x) / cosx limit not exist

Because CON1 / X cannot be taken when Lim x tends to - 0 or + 0

Why does Lim 1-cosx / 1 + cosx (x tends to infinity) have no limit

This limit does not exist because cosx is an oscillatory function

It is proved that the discontinuities of monotone functions on R are at most countable

Let f (x) monotonically increase on R. let the set of discontinuities of F (x) be a. for a ∈ a, define L (a) = Lim {x → a -} f (x), R (a) = Lim {x → a +} f (x). Monotonically increase by f (x), l (a), R (a) exist, and l (a) ≤ f (a) ≤ R (a). While a is a breakpoint, l (a) < R (a), otherwise L

If function f is monotonic on interval I, then any breakpoint of F on I must be the first type of breakpoint. Is it correct?

Let f monotonically increase on interval I
For all a ∈ I, it is only necessary to prove that the left and right limits of F exist at point a
1. Take xn