`lim_( x to 1) {log_e (e^(x(x-1))-e^(x(1-x))) - log_e(4x(x-1))}` is equal to :

Options

A. `-2 log_e 2`
B. `1`
C. `1-log_e 2`
D. `-log_e 2`

Answer

To solve the limit

\[ \lim_{x \to 1} \left( \log_e \left( e^{x(x-1)} - e^{x(1-x)} \right) - \log_e(4x(x-1)) \right), \]

we can start by applying the logarithmic property that states \(\log_a b - \log_a c = \log_a \left( \frac{b}{c} \right)\). Thus, we can rewrite the limit as:

\[ \lim_{x \to 1} \log_e \left( \frac{e^{x(x-1)} - e^{x(1-x)}}{4x(x-1)} \right). \]

Next, we simplify the expression inside the logarithm. Notice that:

\[ e^{x(1-x)} = e^{x - x^2} = e^{x(x-1)}. \]

So we can rewrite the numerator:

\[ e^{x(x-1)} - e^{x(1-x)} = e^{x(x-1)} - e^{-x(x-1)}. \]

This can be expressed using the identity \(a - b = (a - b)\) where \(a = e^{x(x-1)}\) and \(b = e^{-x(x-1)}\):

\[ e^{x(x-1)} - e^{-x(x-1)} = 2 \sinh(x(x-1)). \]

Thus, we can rewrite our limit as:

\[ \lim_{x \to 1} \log_e \left( \frac{2 \sinh(x(x-1))}{4x(x-1)} \right). \]

Now, we know that \(\sinh(t) \approx t\) as \(t \to 0\). Therefore, as \(x \to 1\), \(x(x-1) \to 0\), and we can use the approximation:

\[ \sinh(x(x-1)) \approx x(x-1). \]

Substituting this back into our limit gives:

\[ \lim_{x \to 1} \log_e \left( \frac{2 \cdot x(x-1)}{4x(x-1)} \right) = \lim_{x \to 1} \log_e \left( \frac{2}{4} \right) = \lim_{x \to 1} \log_e \left( \frac{1}{2} \right). \]

Since \(\log_e \left( \frac{1}{2} \right) = -\log_e(2)\), we find that:

\[ \lim_{x \to 1} \left( \log_e \left( e^{x(x-1)} - e^{x(1-x)} \right) - \log_e(4x(x-1)) \right) = -\log_e(2). \]

Thus, the final answer is:

\[ -\log_e(2). \]