Limits, Continuity & Differentiability Question - None

Question

If $\lim_{x \to 0} \left[1 + x + \frac{f(x)}{x}\right]^{1/x} = e^3$, then $\lim_{x \to 0} \frac{f(x)}{x}$ is equal to

Accepted Answer

Step 1: Let $\lim_{x \to 0} \frac{f(x)}{x} = L$. Then $\lim_{x \to 0} \left[1 + x + L\right]^{1/x}$

Step 2: For this to equal $e^3$, use the form $\lim_{x \to 0} [1 + u(x)]^{v(x)} = e^{\lim u(x) \cdot v(x)}$

Step 3: Here $u(x) = x + L$ and $v(x) = \frac{1}{x}$, so $\lim_{x \to 0} (x + L) \cdot \frac{1}{x} = \lim_{x \to 0} \left(1 + \frac{L}{x}\right)$

Step 4: For convergence, we need $\lim_{x \to 0} \left(1 + L\right) \cdot \frac{1}{x} = 3$, which gives $L = 2$

∴ Answer is D.

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