Question

$\displaystyle \lim_{x \to 1}$$\left\{log_{e}\left(e^{x\left(x-1\right)}-e^{x\left(1-x\right)}\right)-log_{e}\left(4x\left(x-1\right)\right)\right\}$ is equal to :

• A. $-2\,log_e\,2$
• B. $1$
• C. $1-\,log_e\,2$
• D. $-\,log_e\,2$

Answer 1

The Correct Option is D

To solve the given limit problem, we need to evaluate the following expression:

$\lim_{x \to 1}\left\{\log_{e}\left(e^{x(x-1)}-e^{x(1-x)}\right)-\log_{e}\left(4x(x-1)\right)\right\}$

1. Start by rewriting the expressions inside the logarithms:

   The expression $e^{x(x-1)}$ simplifies to $e^{x^2-x}$ and $e^{x(1-x)}$ simplifies to $e^{x-x^2}$.

2. Observe the subtraction inside the logarithm:

   $e^{x^2-x} - e^{x-x^2}$ can be rewritten as:

   $e^{x^2-x}(1 - e^{2(x-x^2)})$.

   At $x=1$, both terms seem to converge towards zero.

3. Apply L'Hôpital's Rule for indeterminate forms:

   To solve $\lim_{x \to 1}$ of the above expressions directly, we need to simplify:

   We can use approximation around $x=1$ using Taylor series or expansion around $x=1$.

   Using series expansion, assume $x = 1 + h$ where $h \to 0$.

   Evaluate the terms:

   • $x^2 - x = 1 + 2h + h^2 - (1 + h)$ simplifies to $h^2 + h$.
   • $x - x^2 = 1 + h - (1 + 2h + h^2)$ simplifies to -$h^2 - h$.

   Therefore, $e^{x^2-x} \approx e^{h(h+1)}$ and $e^{x-x^2} \approx e^{-h(h+1)}$.

4. Logarithmic simplification:

   Substitute these into the original expression and see that:

   $e^{x(x-1)} - e^{x(1-x)} \approx e^{h(h+1)}(1 - e^{-2h(h+1)})$

   Now, $\log_e(1 - e^{-2h(h+1)}) \to -\log_e 2$ as $x \to 1$, by simplifying logarithms.

5. Evaluate the logarithmic difference:

   Cancel out similar terms, and the expression reduces to a constant:

   Thus, $-\log_e 2$ is left as the limit value:

   -$\log_e 2$.

Following through the logical steps, the correct answer is -$\log_e 2$.