Question

Let \(f : \mathbb{R} \to \mathbb{R}\) be a differentiable function satisfying \(f^{\prime}(3) + f^{\prime}(2) = 0\). Then \(\lim_{x \to 0}\left(\frac{1 + f(3 + x) - f(3)}{1 + f(2 - x) - f(2)}\right)^{\frac{1}{x}}\) is equal to

• A. \(e\)
• B. \(e^2\)
• C. \(e^{-1}\)
• D. \(1\)

Hint:

The form is \(1^\infty\); take the logarithm and use the derivative definitions of \(f^{\prime}(3)\) and \(f^{\prime}(2)\).

Answer 1

The Correct Option is D

At \(x \to 0\) the base tends to \(\frac{1 + 0}{1 + 0} = 1\), so the limit has the form \(1^\infty\).
Let the limit be \(e^{\ell}\) with \(\ell = \lim_{x\to0}\frac{1}{x}\left[\frac{N - D}{D}\right]\), where \(N = 1 + f(3 + x) - f(3)\), \(D = 1 + f(2 - x) - f(2)\).
\(N - D = [f(3 + x) - f(3)] - [f(2 - x) - f(2)]\).
\(\frac{N - D}{x} \to f^{\prime}(3) + f^{\prime}(2)\) because \(\frac{f(2 - x) - f(2)}{x} \to -f^{\prime}(2)\).
\(D \to 1\), so \(\ell = f^{\prime}(3) + f^{\prime}(2) = 0\).
The limit is \(e^0 = 1\).
Answer: (4) 1