x-& gt； 0, so replace all the ones that can be replaced equally first.

The third formula is the variable upper bound integral, which can't be replaced equivalently, and it will be done by L'H?pital's law in the end.

Because the comparison order is not a limit, the coefficient of the limit is not important, and it is finally represented by a and B.

|sin( 1/x)|？ ≤? 1| x |, the absolute value is removed first, because both sides are square because they are positive numbers.

Left = sin ( 1/x) 2

Right = 1/x 2

Is it to prove that it is true for all real numbers? All I can think of is

Function = sin (1/x) 2-1/x 2? (That's left-right)

And then divided into monotonous intervals (0 should be a discontinuous point),

Then take the derivative of this function to see whether the derivative is positive or negative (it should be possible to find out that the derivative is positive when it is greater than 0 and negative when it is less than 0).

Then from this monotonous interval and derivative, we can see that this function gets the maximum at infinity and infinity.

This maximum value, calculated according to the limit, should all be 0.

So it can be proved that the left

I don't think it is necessary to prove this thing. I'll post the left and right function maps to you.

It's obvious when you look at the picture.

sin( 1/x)^2

/photos/25675806 @ N02/72864 154 12/in/photostream

1/x^2

/photos/25675806 @ N02/72863904 12/in/photostream

Left-right

/photos/25675806 @ N02/7286472088/in/photostream