Math Snap

STEP 1

1. We are dealing with a limit problem involving a product of two functions: $\tan x$ and $\ln x$.
2. The limit is taken as $x$ approaches $0$ from the positive side, denoted by $0^+$.
3. The function $\ln x$ approaches $-\infty$ as $x$ approaches $0^+$, and $\tan x$ approaches $0$.

STEP 2

1. Analyze the behavior of $\tan x$ and $\ln x$ as $x \to 0^+$.
2. Use limit properties or transformations to evaluate the limit.
3. Conclude the result of the limit.

STEP 3

First, analyze the behavior of each function separately as $x \to 0^+$.
- $\tan x$ approaches $0$ as $x$ approaches $0$ from the positive side.
- $\ln x$ approaches $-\infty$ as $x$ approaches $0^+$.

STEP 4

Since the limit involves an indeterminate form $0 \cdot (-\infty)$, we need to transform it into a form suitable for L'Hôpital's Rule or another method.
Rewrite the expression $\tan x \cdot \ln x$ as:
$$\frac{\ln x}{\cot x}$$ Now, as $x \to 0^+$, $\ln x \to -\infty$ and $\cot x \to \infty$, resulting in the indeterminate form $\frac{-\infty}{\infty}$.

STEP 5

Apply L'Hôpital's Rule, which is applicable for indeterminate forms $\frac{\infty}{\infty}$ or $\frac{0}{0}$.
Differentiate the numerator and the denominator:
- The derivative of $\ln x$ is $\frac{1}{x}$.
- The derivative of $\cot x$ is $-\csc^2 x$.
Apply L'Hôpital's Rule:
$$\lim_{x \to 0^+} \frac{\ln x}{\cot x} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\csc^2 x}$$ Simplify:
$$= \lim_{x \to 0^+} \frac{-\sin^2 x}{x}$$

STEP 6

Evaluate the limit:
As $x \to 0^+$, $\sin x \approx x$, so $\sin^2 x \approx x^2$.
$$\lim_{x \to 0^+} \frac{-\sin^2 x}{x} = \lim_{x \to 0^+} \frac{-x^2}{x} = \lim_{x \to 0^+} -x = 0$$

Conclude the result of the limit:
The limit is:
$$\boxed{0}$$