Why does the function f (x) = cosx have a limit when x approaches 0 and the limit value is 1, but when x approaches ∞, its limit does not exist?

Why does the function f (x) = cosx have a limit when x approaches 0 and the limit value is 1, but when x approaches ∞, its limit does not exist?

Because when x approaches 0, the value of the function approach can be determined
When x approaches infinity, you can't determine the value of the function approaching
Because the function is a periodic function on R

A variable whose limit is zero is called infinitesimal, but infinitesimal is not necessarily zero. Why?

You also say that the limit of infinitesimal is 0. This sentence means how close the distance between this quantity and zero is, but it does not reach zero. You can simply understand that infinitesimal is a curve with zero as the asymptote, and zero is the X axis (fixed straight line). The curve may be very slow and very close to the X axis, but it is an asymptote after all. You can't say that it coincides with the X axis. You say it is the X axis (infinitesimal is zero) that's even more wrong

Finding the general solution of the differential equation dy / DX = (1 + x) y

The method of separating variables dy / y = (1 + x) DX, integrating on both sides, get ln|y| = x + x square / 2 + C, and sort out the (x + x square / 2) square of y = CE

Find the differential equation y-dy / DX = 1 + X × General solution of dy / DX

(1/y)dy=[x/(1+x^2)]dx
Both sides integrate at the same time
Ln|y| = 1 / 2ln (1 + x ^ 2) + C1 (C1 is any constant)
So y = (1 + x ^ 2) ^ 1 / 2 + C (C is any constant)

Finding the general solution of the differential equation dy / DX + (1 / x) y = e ^ X / X

Let P = 1 / x, q = e ^ X / X
Directly solve the Bernoulli equation,
y=e^(∫-pdx)(∫Qe^(∫pdx)dx+C)
=(1/x)(∫(e^x/x)xdx+C)
=(1/x)(e^x+C)
So y = (e ^ x + C) / X

Find the general solution of equation ylnydx + (x-lny) dy = 0

dx/dy+1/(ylny)*x=1/y
x=e^(∫-1/(ylny)dy){∫1/y*e^[∫1/(ylny)*dy]dy+C}
=1/lny[∫(-1/y*lny)dy+C]
=1/lny[-1/2*ln^2(y)+C]