Limx tends to 0 [(1 / x) - (1 / SiNx)]

Limx tends to 0 [(1 / x) - (1 / SiNx)]

Limx tends to 0 [(1 / x) - (1 / SiNx)]
=
Limx tends to 0 [(sinx-x) / xsinx)]
=lim(x->0) (sinx-x)/x²
=lim(x->0) (cosx-1)/2x
=lim(x->0) (-sinx)/2
=0

Limx tends to 0 SiNx / X
Limx tends to 0 SiNx / x = 1 why!

L'Hospital Law:lim x- >0 f(x)/g(x)=lim x->0 f'(x)/g'(x)
So Lim X - > 0 SiNx / x = Lim X - > 0 cosx / 1 = cos0 = 1

Limx → 0xsin (1 / x) = 0 limx →∞ xsin (1 / x) = 1 limx →∞ (1 / x) SiNx = 1 why? I think they are all equal to 1

These three are all indefinite integrals,
First: limx → 0, xsin (1 / x) = 0
X is an infinitesimal; sin (1 / x) is equivalent to sin ∞, but is a bounded variable (± 1)
Infinitesimal times bounded variable or infinitesimal, so the limit is 0
Second: limx →∞ xsin (1 / x) = 1
X is an infinitesimal quantity; sin (1 / x) is equivalent to sin0
The result of infinitesimal quantity multiplied by infinitesimal quantity is possible in three cases (0, ∞, constant),
The second can be converted into one of two important limits to continue
Third: limx →∞ (1 / x) SiNx = 1
It's essentially the same as the second one

Limx times 2 / 2 of SiNx, X tends to infinity. Why is 2 / 2 of SiNx equivalent to x
I don't know why it can be replaced, and can we ignore sin cos directly by multiplication and division?
Sin 2 / X is equivalent to 2 / X. please write it wrong

The original formula = limxsin (2 / x)? If so, it is as follows
Let t = 2 / x, then x = 2 / T
When x tends to infinity, t tends to 0, then Sint is equivalent to t
This is called the Equivalent Infinitesimal Substitution. It holds only when t tends to 0. The textbook gives three groups of equivalent infinitesimals. Remember, there is another group where TaNx is equivalent to X when x tends to 0,
You go here and have a look
Original formula = LIM (2 / T) Sint = 2