Given f(x) = √ x^3 + x^2 sin (π/x ) + 2:

Question

(b) Find limx→0 g(x) and limx→0 h(x), and use the Squeeze Theorem to determine limx→0 f(x).

Answer 1

The usual notation uses g(x) ≤ f(x) ≤ h(x).

As -x² ≤ x²sin (π/x ) ≤ x², then

x^3/2 - x² + 2 ≤ x^3/2 + x^2 sin (π/x ) + 2 ≤ x^3/2 + x^2 + 2.

You can chose g and h as,

g(x) = x^3/2 - x² + 2

h(x) = x^3/2 + x² + 2

a) lim_x→0 g(x) = 2

b) lim_x→0 h(x) = 2

c) As lim_x→0 g(x) = lim_x→0 h(x) Squeeze Theorem gives lim_x→0 f(x) = 2