Question

$\lim_{x \to 2024} \sqrt[1]{(x-2023)^{2023}(x-2024)^{2024}\cdot (x-2025)^{2025}}$

Answer 1

0

Answer 2

The given limit is $\lim_{x \to 2024} \sqrt[1]{(x-2023)^{2023}(x-2024)^{2024}\cdot (x-2025)^{2025}}$.

The notation $\sqrt[1]{y}$ represents the first root of $y$, which is $y^{1/1} = y$.
So, the expression simplifies to $(x-2023)^{2023}(x-2024)^{2024}\cdot (x-2025)^{2025}$.

Let $f(x) = (x-2023)^{2023}(x-2024)^{2024}\cdot (x-2025)^{2025}$.
This function is a product of terms of the form $(x-a)^n$, where $n$ is a positive integer. Such terms are polynomials in $x$. The product of polynomials is a polynomial.
Polynomial functions are continuous for all real numbers.
Since $f(x)$ is a polynomial, it is continuous at $x=2024$.
For a continuous function, the limit as $x$ approaches a point is equal to the function value at that point.
Therefore, $\lim_{x \to 2024} f(x) = f(2024)$.

Substitute $x=2024$ into the expression for $f(x)$:
$f(2024) = (2024-2023)^{2023}(2024-2024)^{2024}\cdot (2024-2025)^{2025}$
$f(2024) = (1)^{2023}(0)^{2024}\cdot (-1)^{2025}$

Evaluate each term:
$(1)^{2023} = 1$
$(0)^{2024} = 0$ (since the exponent is positive, $0^n = 0$ for $n > 0$)
$(-1)^{2025} = -1$ (since the exponent is odd, $(-1)^n = -1$ for odd $n$)

Now, multiply the results:
$f(2024) = 1 \cdot 0 \cdot (-1) = 0$.

Thus, the limit of the function as $x \to 2024$ is 0.