Question

Suppose that f(x) is bounded: that is, there exists a constant M such that abs(f(x)) is < or equal to M for all x. Use the squeeze theorem to prove that lime x^2f(x)=0 as x approaches 0.

Answer 1

First prove that:
Lim x²*M = 0 as x->0 and
Lim x²*(-M) = 0 as x->0

Since it is given that f(x)<M if f(x)>0, and f(x)>-M if f(x)<0, it follows by the squeeze theorem that Lim x²f(x)=0 as x->0.

Answer 2

if g(x) is Mx^2 then what is f(x) and h(x) according to the squeeze theorem.

Answer 3

If g(x)=Mx², and h(x)=-Mx²
then
h(x) ≤ f(x) ≤ g(x)

and if
Lim g(x) (x->0) = Lim h(x) (x->0)
then by the squeeze theorem,
Lim g(x) (x->0) = Lim f(x) (x->0) = Lim h(x) (x->0)