Question

Let \(\mathrm{f}(\mathrm{x})\) be defined for all \(\mathrm{x} \in \mathrm{R}\) such that $\lim _{x \rightarrow 0}\left[f(x)+\ln \left(1-\frac{1}{\mathrm{e}^{f(x)}}\right)-\ln (f(x))\right]=0$ then \(\mathrm{f}(0)\) is (A) 0 (B) 1 (C) 2 (D) 3

Short answer

Now that we have found the solution, the short answer is as follows: Given the limit equation, we manipulated it using logarithmic properties, exponentiation, simplification, and L'Hôpital's rule to determine the value of f(0). We found that f(0) = 0, so the correct answer is (A) 0.

Step-by-step solution

Rewrite the given limit with logarithmic properties

First, rewrite the given limit equation with logarithmic properties: \(\lim_{x \rightarrow 0} \left[ f(x) + \ln \left(1-\frac{1}{\mathrm{e}^{f(x)}}\right) - \ln(f(x))\right] = 0\) Using the property: \(\ln(a) - \ln(b) = \ln\left(\frac{a}{b}\right)\), we have: \(\lim_{x \rightarrow 0} \left[ \ln \left(\frac{1-\frac{1}{\mathrm{e}^{f(x)}}}{f(x)}\right) + f(x) \right] = 0\)

Use exponentiation to simplify the limit equation

Next, we will exponentiate both sides of the equation to eliminate the logarithm: \(\lim_{x \rightarrow 0} \left[ \mathrm{e}^{\ln\left(\frac{1-\frac{1}{\mathrm{e}^{f(x)}}}{f(x)}\right)+f(x)}\right] = 1\) Observe that \(e^{\ln(x)} = x\). This simplifies the equation to: \(\lim_{x \rightarrow 0} \left[\frac{1-\frac{1}{\mathrm{e}^{f(x)}}}{f(x)} \cdot \mathrm{e}^{f(x)}\right] = 1\)

Simplify the given equation

To further simplify, we will multiply the numerator and denominator by \(\mathrm{e}^{f(x)}\): \(\lim_{x \rightarrow 0} \left[\frac{1-\frac{1}{\mathrm{e}^{f(x)}}}{f(x)} \cdot \frac{\mathrm{e}^{f(x)}}{\mathrm{e}^{f(x)}}\right] = \mathrm{e}^0\) Therefore, the equation becomes: \(\lim_{x \rightarrow 0} \left[\frac{\mathrm{e}^{f(x)}-1}{f(x)\mathrm{e}^{f(x)}}\right] = 1\)

Apply L'Hôpital's rule to the limit equation

Now apply L'Hôpital's rule on the limit when x approaches 0: \(\lim_{x \rightarrow 0} \left[\frac{\frac{\mathrm{d}}{\mathrm{d}x}\left(\mathrm{e}^{f(x)}-1\right)}{\frac{\mathrm{d}}{\mathrm{d}x}(f(x)\mathrm{e}^{f(x)})}\right] = 1\) Evaluate the derivatives: \(\lim_{x \rightarrow 0} \left[\frac{f'(x)\mathrm{e}^{f(x)}}{f'(x)\mathrm{e}^{f(x)}+f(x)f'(x)\mathrm{e}^{f(x)}}\right] = 1\)

Simplify the limit equation and solve for f(0)

Cancel out common terms in the numerator and denominator: \(\lim_{x \rightarrow 0} \left[\frac{1}{1+f(x)}\right] = 1\) Since the limit exists and equals 1, we have: \(\frac{1}{1+f(0)} = 1\) Solve for f(0): \(1+f(0)=1\) \(f(0)=0\) So, the correct answer is: (A) 0