Solve Continuous Function Homework: Render Graphic & Proof h: [0,1]→R

[porednata]

Homework Statement

Do the graphic rendering (and write the full proof) of the function h : [0,1) -> $$\Re$$ , which is continuous and bounded but does not reach it's bounds.

2. The attempt at a solution
If h is continuous : exists
lim h(x) = h(x0)
x->xo
If h is bounded:
А ≤ h(x) ≤ В for every x$$\in$$[0, 1)

I thought that the function f(x) = $$\frac{sin\frac{1}{1-x}}{2-x}$$
works but I can't quite do the proof and the graphic rendering which leads me to the point that I'm wrong. Pls help
f(x) = $$\frac{sin\frac{1}{1-x}}{2-x}$$ 's graphic rendering :
View attachment untitled.bmp
is this right?!

 

[Nino MMP]

![enter image description here][1]Proof: Let h(x) = \frac{sin\frac{1}{1-x}}{2-x}. We need to prove that h is bounded and continuous on [0,1). First, we prove that h is bounded. Since 0 ≤ sin(x) ≤ 1 for all x ∈ \Re, we have 0 ≤ sin(\frac{1}{1-x}) ≤ 1 for all x ∈ [0,1). Therefore, 0 ≤ \frac{sin\frac{1}{1-x}}{2-x} ≤ \frac{1}{2-x} for all x ∈ [0,1). Since \frac{1}{2-x} is a continuous function on [0,1), it is bounded on [0,1). Hence, h is bounded on [0,1). Now, we prove that h is continuous on [0,1). We use the definition of continuity. Let x0 ∈ [0,1). For any ε > 0, there exists δ > 0 such that |x - x0| < δ ⇒ |h(x) - h(x0)| < ε. We need to find such a δ. Since 0 ≤ sin(x) ≤ 1 for all x ∈ \Re, we have |sin(x) - sin(x0)| ≤ 1 for all x, x0 ∈ \Re. Therefore, |h(x) - h(x0)| = |\frac{sin\frac{1}{1-x}}{2-x} - \frac{sin\frac{1}{1-x0}}{2-x0}| ≤ |\frac{sin\frac{1}{1-x} - sin\frac{1}{1-x0}}{2-x}| ≤ |\frac{1}{2-x}| < ε if |x - x0| < min{δ, |2-x0|ε}. Therefore, the function h