Limits, Continuity & Differentiability Question - None

Question

53. 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 $\displaystyle\lim_{x \to 0} (1 + f(x))^{\frac{1}{x}}$ is equal to:

Accepted Answer

Step 1: Sum the geometric series

Given: f'(x) + (f'(x))² + (f'(x))³ + ... = e^x

This is a geometric series with first term a = f'(x) and common ratio r = f'(x).

For |f'(x)| < 1, the sum is: 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) + e^x·f'(x) = e^x

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

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

Step 3: Integrate to find f(x)

f(x) = ∫ e^x/(1 + e^x) dx

Let u = 1 + e^x, then du = e^x dx

f(x) = ∫ du/u = ln|u| + C = ln(1 + e^x) + C

Step 4: Use initial condition f(0) = 0

0 = ln(1 + e^0) + C = ln(2) + C

C = -ln(2)

Therefore: f(x) = ln(1 + e^x) - ln(2) = ln((1 + e^x)/2)

Step 5: Evaluate the limit

L = lim_{x→0} (1 + f(x))^(1/x)

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

This is of form 1^∞. Using ln(L) = lim_{x→0} [ln(1 + f(x))]/x:

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

At x = 0: f(0) = 0, so the numerator → ln(1) = 0 (indeterminate 0/0)

Step 6: Apply L'Hôpital's Rule

ln(L) = lim_{x→0} [f'(x)/(1 + f(x))]/1

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

At x = 0: numerator = 1/2, denominator = 1 + ln(1) = 1

ln(L) = 1/2

L = e^(1/2) = √e

∴ Answer: B

Question:

Step-by-Step Solution

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