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Math  /  Calculus

Questionlim⁡x→0+(ln⁡x−1x)\lim _{x \rightarrow 0^{+}}\left(\ln x-\frac{1}{x}\right)

Studdy Solution

STEP 1

1. The problem involves finding the limit of the expression ln⁡x−1x\ln x - \frac{1}{x} as xx approaches 00 from the right.
2. We need to analyze the behavior of both ln⁡x\ln x and −1x-\frac{1}{x} separately as xx tends to 0+0^+.
3. The natural logarithm function ln⁡x\ln x is defined for x>0x > 0 and approaches −∞-\infty as xx approaches 0+0^+.
4. The function −1x-\frac{1}{x} diverges to −∞-\infty as xx approaches 0+0^+.

STEP 2

1. Analyze the behavior of ln⁡x\ln x as x→0+x \rightarrow 0^+.
2. Analyze the behavior of −1x-\frac{1}{x} as x→0+x \rightarrow 0^+.
3. Determine the combined behavior of ln⁡x−1x\ln x - \frac{1}{x} as x→0+x \rightarrow 0^+.
4. Conclude the limit.

STEP 3

Analyze the behavior of ln⁡x\ln x as x→0+x \rightarrow 0^+.
As xx approaches 0+0^+, ln⁡x\ln x approaches −∞-\infty. lim⁡x→0+ln⁡x=−∞ \lim_{x \rightarrow 0^+} \ln x = -\infty

STEP 4

Analyze the behavior of −1x-\frac{1}{x} as x→0+x \rightarrow 0^+.
As xx approaches 0+0^+, −1x-\frac{1}{x} also approaches −∞-\infty. lim⁡x→0+−1x=−∞ \lim_{x \rightarrow 0^+} -\frac{1}{x} = -\infty

STEP 5

Determine the combined behavior of ln⁡x−1x\ln x - \frac{1}{x} as x→0+x \rightarrow 0^+.
Since both ln⁡x→−∞\ln x \rightarrow -\infty and −1x→−∞-\frac{1}{x} \rightarrow -\infty, we need to consider their rates of divergence. We observe that −1x-\frac{1}{x} diverges to −∞-\infty much faster than ln⁡x\ln x because −1x-\frac{1}{x} has a pole at x=0x = 0.

STEP 6

Conclude the limit based on the dominant term.
Since −1x-\frac{1}{x} dominates ln⁡x\ln x as x→0+x \rightarrow 0^+, the expression ln⁡x−1x\ln x - \frac{1}{x} will be dominated by −1x-\frac{1}{x}.
Therefore, lim⁡x→0+(ln⁡x−1x)=−∞ \lim_{x \rightarrow 0^+} \left( \ln x - \frac{1}{x} \right) = -\infty

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