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.
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.
if g(x) is Mx^2 then what is f(x) and h(x) according to the squeeze theorem.
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)
To prove that lim x^2f(x) = 0 as x approaches 0 using the squeeze theorem, we need to show that there exist two functions g(x) and h(x), such that g(x) ≤ x^2f(x) ≤ h(x) and both g(x) and h(x) approach 0 as x approaches 0.
Given that f(x) is bounded, we can say that there exists a constant M > 0 such that |f(x)| ≤ M for all x. Now, we can use this inequality to find an upper bound for x^2f(x).
Since |f(x)| ≤ M for all x, we can multiply both sides of this inequality by x^2 (which is nonnegative) to get:
x^2|f(x)| ≤ Mx^2
Now, let's define g(x) = -Mx^2 and h(x) = Mx^2, both of which approach 0 as x approaches 0.
For x ≠ 0, we can rewrite the inequality as:
-g(x) ≤ x^2f(x) ≤ h(x)
Since both -g(x) and h(x) approach 0 as x approaches 0, by the squeeze theorem, we can conclude that x^2f(x) also approaches 0 as x approaches 0.
Therefore, lim x^2f(x) = 0 as x approaches 0.