Limits, Continuity & Differentiability Question - None

Question

Let $$f(x)$$ be a real valued function defined for all $$x eq -1, 1$$, satisfying $$f(1) = 1$$ and $$f'(x) = \frac{1}{x^2 + (f(x))^2}$$; then $$\lim_{x o \infty} f(x)$$

Accepted Answer

Step 1: Analyze the derivative f'(x) = 1/(x² + (f(x))²)

Since x² + (f(x))² > 0 for all x in the domain, we have f'(x) > 0 everywhere. This means f(x) is strictly increasing.

Step 2: Establish an upper bound for f'(x)

For any x and f(x), we have: f'(x) = 1/(x² + (f(x))²) ≤ 1/x² (since (f(x))² ≥ 0)

Step 3: Integrate to find convergence

Since f is increasing and f'(x) ≤ 1/x², we can write:

f(x) - f(1) = ∫₁ˣ f'(t) dt ≤ ∫₁ˣ 1/t² dt = [-1/t]₁ˣ = 1 - 1/x < 1

Therefore: f(x) ≤ f(1) + 1 = 1 + 1 = 2 for all x ≥ 1

Step 4: Apply Monotone Convergence Theorem

Since f(x) is strictly increasing and bounded above by 2, the limit lim_{x→∞} f(x) exists and is finite.

Step 5: Determine the exact bound

From the integral estimate: f(x) - 1 ≤ ∫₁ˣ 1/t² dt = 1 - 1/x

Therefore: lim_{x→∞} f(x) ≤ 1 + 1 = 2

More precisely, since f'(x) = 1/(x² + (f(x))²) and f is bounded, as x → ∞:

∫₁^∞ f'(x) dx = lim_{x→∞} f(x) - f(1) converges and equals lim_{x→∞} f(x) - 1

This limit is bounded: lim_{x→∞} f(x) - 1 < ∫₁^∞ 1/t² dt = 1

So: lim_{x→∞} f(x) < 1 + 1 = 2

The precise calculation gives: lim_{x→∞} f(x) exists and is less than 1 + π/4 (which follows from a more refined analysis using the relation with arctan)

∴ Answer: Not specified

Question:

Step-by-Step Solution

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