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Math  /  Calculus

Questionlim⁡x→0+(tan⁡x⋅ln⁡x)\lim _{x \rightarrow 0^{+}}(\tan x \cdot \ln x)

Studdy Solution

STEP 1

1. We are dealing with a limit problem involving a product of two functions: tan⁡x\tan x and ln⁡x\ln x.
2. The limit is taken as xx approaches 00 from the positive side, denoted by 0+0^+.
3. The function ln⁡x\ln x approaches −∞-\infty as xx approaches 0+0^+, and tan⁡x\tan x approaches 00.

STEP 2

1. Analyze the behavior of tan⁡x\tan x and ln⁡x\ln x as x→0+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→0+x \to 0^+.
- tan⁡x\tan x approaches 00 as xx approaches 00 from the positive side. - ln⁡x\ln x approaches −∞-\infty as xx approaches 0+0^+.

STEP 4

Since the limit involves an indeterminate form 0⋅(−∞)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⋅ln⁡x\tan x \cdot \ln x as:
ln⁡xcot⁡x\frac{\ln x}{\cot x}
Now, as x→0+x \to 0^+, ln⁡x→−∞\ln x \to -\infty and cot⁡x→∞\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 00\frac{0}{0}.
Differentiate the numerator and the denominator:
- The derivative of ln⁡x\ln x is 1x\frac{1}{x}. - The derivative of cot⁡x\cot x is −csc⁡2x-\csc^2 x.
Apply L'Hôpital's Rule:
lim⁡x→0+ln⁡xcot⁡x=lim⁡x→0+1x−csc⁡2x\lim_{x \to 0^+} \frac{\ln x}{\cot x} = \lim_{x \to 0^+} \frac{\frac{1}{x}}{-\csc^2 x}
Simplify:
=lim⁡x→0+−sin⁡2xx= \lim_{x \to 0^+} \frac{-\sin^2 x}{x}

STEP 6

Evaluate the limit:
As x→0+x \to 0^+, sin⁡x≈x\sin x \approx x, so sin⁡2x≈x2\sin^2 x \approx x^2.
lim⁡x→0+−sin⁡2xx=lim⁡x→0+−x2x=lim⁡x→0+−x=0\lim_{x \to 0^+} \frac{-\sin^2 x}{x} = \lim_{x \to 0^+} \frac{-x^2}{x} = \lim_{x \to 0^+} -x = 0

STEP 7

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

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