Questions about function limits

1.n=2k

The limit is ∞

n=2k+1

The limit is Lópida’s rule

lim sin(narccosx)*n*1/sqrt(1-x^2)=nsin(nπ/2)

2.n=4k+1, the limit is n

3.n=4k+3, the limit is -n

Never learned it?

You should know that lim sinx/x=1, lim arcsinx/x=1, otherwise you can’t do it

Even so, the intermediate process is complicated.

Lópida's rule is the best

Infinitely small/infinitely small or infinitely large/infinitely large

You can use upper and lower derivation respectively,

If the latter exists or is infinite, it is equal to the former; if the latter does not exist, there is no guarantee that the former must exist.

The equivalent infinitesimal substitution will always be, for example, limA/limB=1, then it can be in the limit

lim CA=lim CA/B*B=lim CB*lim (A /B)=lim CB

Let me tell you about my ideas. I wrote down a whole bunch of them.

For example, 1.n=4k+1

cos(narccosx)=sin(nπ/2-narccosx)

and nπ/2-narccosx are equal The valence is infinitesimal, replace it with nπ/2-narccosx

It becomes lim (nπ/2-narccosx)/x

And arcsinx+arccosx=π/2, so

lim (nπ/2-narccosx)/x=lim narcsinx/x=n

Another truth

2.n=4k+3

cos(narccosx)=-sin(nπ/2-narccosx)

The only difference from the above is that there is a negative sign here, it is the same below