Limits, Continuity & Differentiability Question - None

Question

If $ f'(x) + (f'(x))^2 + (f'(x))^3 + (f'(x))^4 + \cdots \infty = e^x $, where $ f'(x) \in (-1, 1) $ and $ f(0) = 0 $, then the value of $ \lim_{x \to 0} (1 + f(x))^{\frac{1}{x}} $ is equal to:

Accepted Answer

Step 1: Sum the infinite geometric series on the left side.

For |f'(x)| < 1, the series f'(x) + (f'(x))² + (f'(x))³ + ... = f'(x)/(1 - f'(x))

So: f'(x)/(1 - f'(x)) = e^x

Step 2: Solve for f'(x).

f'(x) = e^x(1 - f'(x))

f'(x) = e^x - e^x·f'(x)

f'(x)(1 + e^x) = e^x

f'(x) = e^x/(1 + e^x)

Step 3: Integrate to find f(x) using f(0) = 0.

f(x) = ∫₀ˣ e^t/(1 + e^t) dt

Let u = 1 + e^t, then du = e^t dt

f(x) = ∫₁^(1+e^x) du/u = ln(1 + e^x) - ln(2) = ln((1 + e^x)/2)

Step 4: Evaluate the limit.

lim_{x→0} (1 + f(x))^(1/x) = lim_{x→0} (1 + ln((1 + e^x)/2))^(1/x)

At x = 0: (1 + f(0))^(1/0) is indeterminate since f(0) = ln(1) = 0 and we have 1^∞

Use: L = e^(lim_{x→0} ln(1 + f(x))/x)

Since f(x) = ln((1 + e^x)/2) and f'(0) = e^0/(1 + e^0) = 1/2

lim_{x→0} f(x)/x = f'(0) = 1/2 (by L'Hôpital or definition)

Therefore: lim_{x→0} ln(1 + f(x))/x = lim_{x→0} ln(1 + f(x))·(1/f(x))·(f(x)/x) = 1·1·(1/2) = 1/2

∴ L = e^(1/2) = √e

Question:

Step-by-Step Solution

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