Question

$\lim_{x \to 0} \left( \frac{(e^x-e^{-x}) \sin(x)}{20^x21 - 21^x20 + 1} \right)^{(e^{x-x})}$ is equal to

Answer 1

0

Answer 2

To evaluate the limit $\lim_{x \to 0} \left( \frac{(e^x-e^{-x}) \sin(x)}{20^x21 - 21^x20 + 1} \right)^{(e^{x-x})}$, we follow these steps:

1. Simplify the exponent: The exponent is $e^{x-x}$. $e^{x-x} = e^0 = 1$. So the expression simplifies to: $\lim_{x \to 0} \left( \frac{(e^x-e^{-x}) \sin(x)}{20^x21 - 21^x20 + 1} \right)^{1} = \lim_{x \to 0} \frac{(e^x-e^{-x}) \sin(x)}{20^x21 - 21^x20 + 1}$

2. Evaluate the limit of the numerator: Let the numerator be $N(x) = (e^x-e^{-x}) \sin(x)$. Substitute $x=0$ into $N(x)$: $\lim_{x \to 0} N(x) = (e^0 - e^{-0}) \sin(0) = (1 - 1) \cdot 0 = 0 \cdot 0 = 0$

3. Evaluate the limit of the denominator: Let the denominator be $D(x) = 20^x21 - 21^x20 + 1$. Substitute $x=0$ into $D(x)$: $\lim_{x \to 0} D(x) = 20^0 \cdot 21 - 21^0 \cdot 20 + 1 = 1 \cdot 21 - 1 \cdot 20 + 1 = 21 - 20 + 1 = 2$

4. Combine the limits: Since the limit of the numerator is $0$ and the limit of the denominator is $2$ (which is a non-zero constant), the limit of the fraction is: $\lim_{x \to 0} \frac{N(x)}{D(x)} = \frac{\lim_{x \to 0} N(x)}{\lim_{x \to 0} D(x)} = \frac{0}{2} = 0$

Thus, the value of the given limit is $0$.

Explanation: The exponent $e^{x-x}$ simplifies to $e^0 = 1$. The problem then reduces to evaluating the limit of a rational function as $x \to 0$. The numerator $(e^x-e^{-x})\sin(x)$ approaches $(e^0-e^0)\sin(0) = (1-1)\cdot 0 = 0$. The denominator $20^x21 - 21^x20 + 1$ approaches $20^0 \cdot 21 - 21^0 \cdot 20 + 1 = 1 \cdot 21 - 1 \cdot 20 + 1 = 2$. Since the numerator approaches 0 and the denominator approaches a non-zero value (2), the limit of the entire expression is $\frac{0}{2} = 0$.