Trig Review Problem

I think that the  inequality  is incorrect

See the graph here :  https://www.desmos.com/calculator/alijngoclf

Notice that x^2   is  ≥ sin^2x    everywhere on   (-pi/2, pi/2)

But  if     x^2  ≤ f(x)   on this interval...then f(x) is ≥ x^2  everywhere on this interval...then f(x)  cannot also be  ≤ sin^2x  everywhere on the interval

I believe that the inequality we need is  this

sin^2 x  ≤ f(x)   ≤ x^2

Note that if this true......then

f(x)                 f(x)      =    0          by the Squeeze Theorem

lim x → 0