Math Problem Statement

Given that 1 - (x^2 / 4) ≤ u(x) ≤ 1 + (x^2 / 2) for all x ≠ 0, find lim(x→0) u(x), no matter how complicated u is.

Solution

The given problem is:

1x24u(x)1+x22, for all x0,1 - \frac{x^2}{4} \leq u(x) \leq 1 + \frac{x^2}{2}, \text{ for all } x \neq 0, and we are asked to find:

limx0u(x).\lim_{{x \to 0}} u(x).

Applying the Squeeze Theorem

To find the limit, we can apply the Squeeze Theorem (also known as the Sandwich Theorem). The Squeeze Theorem states that if f(x)g(x)h(x)f(x) \leq g(x) \leq h(x) for all xx in some interval around aa, except possibly at aa itself, and if

limxaf(x)=limxah(x)=L,\lim_{{x \to a}} f(x) = \lim_{{x \to a}} h(x) = L, then

limxag(x)=L.\lim_{{x \to a}} g(x) = L.

In our case:

f(x)=1x24,g(x)=u(x),h(x)=1+x22.f(x) = 1 - \frac{x^2}{4}, \quad g(x) = u(x), \quad h(x) = 1 + \frac{x^2}{2}.

Now, let's find the limits of f(x)f(x) and h(x)h(x) as x0x \to 0.

limx0(1x24)=10=1,\lim_{{x \to 0}} \left(1 - \frac{x^2}{4}\right) = 1 - 0 = 1, limx0(1+x22)=1+0=1.\lim_{{x \to 0}} \left(1 + \frac{x^2}{2}\right) = 1 + 0 = 1.

Since both limits are equal to 1, we can apply the Squeeze Theorem:

limx0u(x)=1.\lim_{{x \to 0}} u(x) = 1.

Conclusion

The limit of u(x)u(x) as x0x \to 0 is:

limx0u(x)=1.\lim_{{x \to 0}} u(x) = 1.

Would you like more details on this solution or have any questions?

Related Questions to Explore:

  1. How does the Squeeze Theorem work for trigonometric functions?
  2. What are some other applications of the Squeeze Theorem?
  3. Can the Squeeze Theorem be used for complex functions?
  4. What happens if one of the bounds diverges to infinity?
  5. How would this problem change if we approached a different point?

Tip:

When using the Squeeze Theorem, ensure that both bounding functions have the same limit at the point of interest!

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Math Problem Analysis

Mathematical Concepts

Limits
Squeeze Theorem
Inequalities

Formulas

Squeeze Theorem: If f(x) ≤ g(x) ≤ h(x) and lim(x→a) f(x) = lim(x→a) h(x) = L, then lim(x→a) g(x) = L
Limit of f(x) = 1 - (x^2 / 4) as x → 0 is 1
Limit of h(x) = 1 + (x^2 / 2) as x → 0 is 1

Theorems

Squeeze Theorem

Suitable Grade Level

Grades 11-12 or College Level