Question

$\lim_{x\to 0}\frac{sinln \frac{2024}{2024-x}}{tan(\frac{2024}{2024-3x}-1)}$

Answer 1

1/3

Answer 2

The given limit is of the indeterminate form $\frac{0}{0}$. We can use standard limits or L'Hopital's rule.

We use the standard limits: $\lim_{u\to 0} \frac{\sin u}{u} = 1$, $\lim_{u\to 0} \frac{\tan u}{u} = 1$, and $\lim_{u\to 0} \frac{\ln(1+u)}{u} = 1$.

Let's analyze the numerator: $\sin\left(\ln \frac{2024}{2024-x}\right)$.

$\ln \frac{2024}{2024-x} = \ln \left(\frac{2024-x+x}{2024-x}\right) = \ln \left(1 + \frac{x}{2024-x}\right)$.

As $x \to 0$, $\frac{x}{2024-x} \to 0$.

Let $u(x) = \ln \left(1 + \frac{x}{2024-x}\right)$. As $x \to 0$, $u(x) \to 0$.

The numerator is $\sin(u(x))$. We can write this as $\frac{\sin(u(x))}{u(x)} \cdot u(x)$.

$\lim_{x\to 0} \frac{\sin(u(x))}{u(x)} = 1$.

Now consider $u(x) = \ln \left(1 + \frac{x}{2024-x}\right)$. Let $v(x) = \frac{x}{2024-x}$. As $x \to 0$, $v(x) \to 0$.

$u(x) = \ln(1+v(x))$. We can write this as $\frac{\ln(1+v(x))}{v(x)} \cdot v(x)$.

$\lim_{x\to 0} \frac{\ln(1+v(x))}{v(x)} = 1$.

So, the numerator is asymptotically equivalent to $v(x) = \frac{x}{2024-x}$ as $x \to 0$.

$\lim_{x\to 0} \frac{\sin\left(\ln \frac{2024}{2024-x}\right)}{x} = \lim_{x\to 0} \frac{\sin(u(x))}{u(x)} \cdot \frac{\ln(1+v(x))}{v(x)} \cdot \frac{v(x)}{x} = 1 \cdot 1 \cdot \lim_{x\to 0} \frac{\frac{x}{2024-x}}{x} = \lim_{x\to 0} \frac{1}{2024-x} = \frac{1}{2024}$.

Let's analyze the denominator: $\tan\left(\frac{2024}{2024-3x}-1\right)$.

$\frac{2024}{2024-3x}-1 = \frac{2024 - (2024-3x)}{2024-3x} = \frac{3x}{2024-3x}$.

As $x \to 0$, $\frac{3x}{2024-3x} \to 0$.

Let $w(x) = \frac{3x}{2024-3x}$. As $x \to 0$, $w(x) \to 0$.

The denominator is $\tan(w(x))$. We can write this as $\frac{\tan(w(x))}{w(x)} \cdot w(x)$.

$\lim_{x\to 0} \frac{\tan(w(x))}{w(x)} = 1$.

So, the denominator is asymptotically equivalent to $w(x) = \frac{3x}{2024-3x}$ as $x \to 0$.

$\lim_{x\to 0} \frac{\tan\left(\frac{2024}{2024-3x}-1\right)}{x} = \lim_{x\to 0} \frac{\tan(w(x))}{w(x)} \cdot \frac{w(x)}{x} = 1 \cdot \lim_{x\to 0} \frac{\frac{3x}{2024-3x}}{x} = \lim_{x\to 0} \frac{3}{2024-3x} = \frac{3}{2024}$.

Now, we can evaluate the limit of the ratio:

$\lim_{x\to 0}\frac{\sin\left(\ln \frac{2024}{2024-x}\right)}{\tan\left(\frac{2024}{2024-3x}-1\right)} = \lim_{x\to 0}\frac{\frac{\sin\left(\ln \frac{2024}{2024-x}\right)}{x}}{\frac{\tan\left(\frac{2024}{2024-3x}-1\right)}{x}}$.

Using the limits we found for the numerator and denominator divided by $x$:

$= \frac{\lim_{x\to 0}\frac{\sin\left(\ln \frac{2024}{2024-x}\right)}{x}}{\lim_{x\to 0}\frac{\tan\left(\frac{2024}{2024-3x}-1\right)}{x}} = \frac{1/2024}{3/2024} = \frac{1}{3}$.