Given: \(f(x) - f(y) \geq \ln x - \ln y + x - y\)
Rewriting:
\(f(x) - f(y) = \frac{\sqrt{x} - \sqrt{y}}{x - y} \cdot \frac{\sqrt{x} + \sqrt{y}}{x - y}\)
- Case 1: Let \( x \to y \)
\(\lim_{y \to x^-} f' \left( \frac{1}{x} \right) \geq \frac{1}{x} + 1 \quad \text{...(1)}\)
- Case 2: Let \( x \neq y \)
\(\lim_{y \to x^+} f' \left( x^+ \right) \leq \frac{1}{x} + 1 \quad \text{...(2)}\)
Thus, \(f'(x) = \frac{1}{x+1}\).
Now, substitute \( f' \left( \frac{1}{x} \right) = n + 1 \) into the sum:
\(\sum_{n=1}^{20} f' \left( \frac{1}{x} \right) = \sum_{n=1}^{20} \left( n^2 + 1 \right)\)
Calculating:
\(\sum_{n=1}^{20} n^2 = \frac{20 \cdot 21 \cdot 41}{6} = 2870, \quad \sum_{n=1}^{20} 1 = 20\)
Therefore:
\(\sum_{n=1}^{20} f' \left( \frac{1}{x} \right) = 2870 + 20 = 2890\)
\[ \lim_{x \to -\frac{3}{2}} \frac{(4x^2 - 6x)(4x^2 + 6x + 9)}{\sqrt{2x - \sqrt{3}}} \]
\[ f(x) = \begin{cases} \frac{(4^x - 1)^4 \cot(x \log 4)}{\sin(x \log 4) \log(1 + x^2 \log 4)}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \]
Find \( e^k \) if \( f(x) \) is continuous at \( x = 0 \).