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Evaluating a limit that leads to 0/0 by direct substitution

Suppose we know that lim⁡x→3ln⁡(x)=ln⁡(3)\lim_{x \to 3} \ln(x) = \ln(3) and that the function ln⁡(x)\ln(x) is continuous at x=3x=3.

Consider c(x)=ln⁡(x)−ln⁡(3)x−3.c(x) = \frac{\ln(x) - \ln(3)}{x - 3}. A student attempts to evaluate lim⁡x→3c(x)\lim_{x \to 3} c(x) by directly substituting x=3x = 3 into the formula for c(x)c(x). What is the resulting expression, and what type of indeterminate form is it? State your conclusion about whether direct substitution (using the quotient rule for limits) can be used here to find the value of the limit.

Hint

Apply the quotient rule for limits, which requires evaluating the limit of the numerator and denominator separately using continuity/direct substitution. Check whether the quotient rule's hypothesis (denominator limit nonzero) is satisfied.

Step-by-step solution

Since ln⁡(x)\ln(x) is continuous at x=3x=3, direct substitution gives lim⁡x→3(ln⁡(x)−ln⁡(3))=ln⁡(3)−ln⁡(3)=0\lim_{x\to 3}(\ln(x)-\ln(3)) = \ln(3)-\ln(3) = 0. Also, lim⁡x→3(x−3)=3−3=0\lim_{x\to 3}(x-3) = 3-3 = 0 by the difference rule.

The quotient rule for limits requires the limit of the denominator to be nonzero, but here both numerator and denominator limits are 00. So substituting x=3x=3 produces the expression 0/00/0, an indeterminate form, meaning the quotient rule cannot be applied directly to evaluate this limit.

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