Squeeze Theorem Calculus

The question is definitely too long and difficult to explain, I will post an imgur link if you could please access it.

http://imgur.com/TAJYWja

Thank you to anyone who can help, even if it's just one part of it.

Filling in the blanks:
(i) x^2 (1 - 4x^2) ≥ 0 if and only if 1 - 4x^2 ≥ 0.

(ii) This means that x^2 ≥ 4x^4 if and only if x is in [-1/2, 1/2].

(iii) Particularly, we get that for all real numbers x in [-1/2, 1/2], we have f(x) = 1 + 4x^4 ≤ 1 + x^2.

(iv) Therefore, for all x in [-1/2, 1/2], we have that 1 ≤ f(x) ≤ 1 + x^2.

Since lim(x→0) (1 + x^2) = 1, by the Squeeze Theorem, lim(x→0) f(x) = 1.
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I hope this helps!

There's a lot of unnecessary heavy weather in that derivation, but it looks like you might need the practice. The bit about where x^2 - 4x^4 = x^2(1 - 4x^2) >= 0 is pretty simple. That can only be false if x^2 is nonzero and (1-4x^2) is negative. In other words, it's true iff:

1 - 4x^2 >= 0
1 >= 4x^2
1 >= |2x|
-1/2 <= x <= 1/2

That's all elementary algebra stuff. in the interval [-1/2, 1/2] you have
x^2 - 4x^4 >= 0
x^2 >= 4x^4
1 + x^2 >= 1 + 4x^4

Since f(x) is one of those two values, and 1+x^2 is the maximum of the two, then:

f(x) <= 1 + x^2 .... for all x in [-1/2, 1/2]

Combining that with 1 <= f(x) completes the squeeze:

1 <= f(x) <= 1 + x^2

Since the outside expressions both tend to 1 as x-->0, then f(x) must also approach 1 as a limit.