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Evaluate an infinity/infinity limit using L'Hopital's Rule

Evaluate the limit:

lim⁡x→0+ln⁡(x)cot⁡(x)\lim_{x \to 0^+} \frac{\ln(x)}{\cot(x)}

First confirm that direct substitution produces an indeterminate form of type ∞/∞\infty/\infty, then apply L'Hopital's Rule to find the value of the limit.

Hint

Differentiate the numerator and denominator separately (do not use the quotient rule). Recall that ddx[cot⁡(x)]=−csc⁡2(x)\frac{d}{dx}[\cot(x)] = -\csc^2(x), and rewrite −csc⁡2(x)-\csc^2(x) in terms of sin⁡(x)\sin(x) to simplify the resulting limit.

Step-by-step solution

As x→0+x \to 0^+, ln⁡(x)→−∞\ln(x) \to -\infty and cot⁡(x)→+∞\cot(x) \to +\infty, so the limit has the form −∞/∞-\infty/\infty, which is indeterminate and justifies applying L'Hopital's Rule. Differentiating numerator and denominator separately gives ddx[ln⁡(x)]=1x\frac{d}{dx}[\ln(x)] = \frac{1}{x} and ddx[cot⁡(x)]=−csc⁡2(x)\frac{d}{dx}[\cot(x)] = -\csc^2(x).

The limit becomes lim⁡x→0+1/x−csc⁡2(x)=lim⁡x→0+−sin⁡2(x)x\lim_{x\to 0^+} \dfrac{1/x}{-\csc^2(x)} = \lim_{x\to 0^+} \dfrac{-\sin^2(x)}{x}. Since sin⁡2(x)=sin⁡(x)⋅sin⁡(x)\sin^2(x) = \sin(x)\cdot\sin(x) and lim⁡x→0+sin⁡(x)x=1\lim_{x\to 0^+}\frac{\sin(x)}{x} = 1, we can write this as lim⁡x→0+[−sin⁡(x)x⋅sin⁡(x)]=−(1)(0)=0\lim_{x\to 0^+}\left[-\frac{\sin(x)}{x}\cdot \sin(x)\right] = -(1)(0) = 0.

Thus the original limit evaluates to 00.

Sign of the limit value

Hint
Step-by-step solution

The rewritten limit is lim⁡x→0+−sin⁡2(x)x\lim_{x\to 0^+} \dfrac{-\sin^2(x)}{x}. Although the expression −sin⁡2(x)x\dfrac{-\sin^2(x)}{x} is negative for all x>0x > 0 near 00, the limit value itself is 00. Therefore, the correct answer is zero.

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