Question

$\lim_{x \to 1} \left( \frac{2024}{x^{2024}-1} - \frac{2022}{x^{2022}-1} \right)$

• A. 1
• B. -1
• C. 0
• D. The limit does not exist.

Answer 1

Answer: (B)

-1

Answer 2

To solve the limit $\lim_{x \to 1} \left( \frac{2024}{x^{2024}-1} - \frac{2022}{x^{2022}-1} \right)$, we first observe that as $x \to 1$, both denominators $x^{2024}-1$ and $x^{2022}-1$ approach 0. This results in an indeterminate form of type $\infty - \infty$.

We combine the fractions: $L = \lim_{x \to 1} \frac{2024(x^{2022}-1) - 2022(x^{2024}-1)}{(x^{2024}-1)(x^{2022}-1)}$ Let the numerator be $N(x) = 2024(x^{2022}-1) - 2022(x^{2024}-1) = 2024x^{2022} - 2024 - 2022x^{2024} + 2022 = -2022x^{2024} + 2024x^{2022} - 2$. Let the denominator be $D(x) = (x^{2024}-1)(x^{2022}-1)$.

As $x \to 1$, $N(1) = -2022(1)^{2024} + 2024(1)^{2022} - 2 = -2022 + 2024 - 2 = 0$. And $D(1) = (1^{2024}-1)(1^{2022}-1) = (1-1)(1-1) = 0$. This is an indeterminate form of type $\frac{0}{0}$, so we can apply L'Hopital's Rule.

Let $n_1 = 2024$ and $n_2 = 2022$. The expression becomes: $L = \lim_{x \to 1} \frac{n_1(x^{n_2}-1) - n_2(x^{n_1}-1)}{(x^{n_1}-1)(x^{n_2}-1)}$ Numerator: $N(x) = n_1 x^{n_2} - n_1 - n_2 x^{n_1} + n_2$. Denominator: $D(x) = x^{n_1+n_2} - x^{n_1} - x^{n_2} + 1$.

First derivative of $N(x)$: $N'(x) = n_1 n_2 x^{n_2-1} - n_2 n_1 x^{n_1-1}$. $N'(1) = n_1 n_2 - n_2 n_1 = 0$. First derivative of $D(x)$: $D'(x) = (n_1+n_2)x^{n_1+n_2-1} - n_1 x^{n_1-1} - n_2 x^{n_2-1}$. $D'(1) = (n_1+n_2) - n_1 - n_2 = 0$.

Since the first derivatives are also zero at $x=1$, we apply L'Hopital's Rule again. Second derivative of $N(x)$: $N''(x) = n_1 n_2 (n_2-1) x^{n_2-2} - n_2 n_1 (n_1-1) x^{n_1-2}$. $N''(1) = n_1 n_2 (n_2-1) - n_2 n_1 (n_1-1) = n_1 n_2 [(n_2-1) - (n_1-1)] = n_1 n_2 (n_2 - n_1)$.

Second derivative of $D(x)$: $D''(x) = (n_1+n_2)(n_1+n_2-1)x^{n_1+n_2-2} - n_1(n_1-1)x^{n_1-2} - n_2(n_2-1)x^{n_2-2}$. $D''(1) = (n_1+n_2)(n_1+n_2-1) - n_1(n_1-1) - n_2(n_2-1)$.

Now, substitute $n_1 = 2024$ and $n_2 = 2022$: $N''(1) = 2024 \cdot 2022 (2022 - 2024) = 2024 \cdot 2022 \cdot (-2)$.

$D''(1) = (2024+2022)(2024+2022-1) - 2024(2024-1) - 2022(2022-1)$ $D''(1) = (4046)(4045) - 2024(2023) - 2022(2021)$ $D''(1) = (2024+2022)(2024+2021) - 2024(2023) - 2022(2021)$ $D''(1) = (2024^2 + 2024 \cdot 2021 + 2022 \cdot 2024 + 2022 \cdot 2021) - 2024 \cdot 2023 - 2022 \cdot 2021$ $D''(1) = 2024^2 + 2024 \cdot 2021 + 2022 \cdot 2024 - 2024 \cdot 2023$ $D''(1) = 2024(2024 + 2021 - 2023) + 2022 \cdot 2024$ $D''(1) = 2024(4045 - 2023) + 2022 \cdot 2024$ $D''(1) = 2024(2022) + 2022 \cdot 2024 = 2 \cdot 2024 \cdot 2022$.

By L'Hopital's Rule, the limit is: $L = \frac{N''(1)}{D''(1)} = \frac{2024 \cdot 2022 \cdot (-2)}{2 \cdot 2024 \cdot 2022} = -1$