Question

Evaluate: \[ \lim_{x\to e}\frac{\log x-1}{x-e} \]

• A. \(1\)
• B. \(\frac12\)
• C. \(\frac1e\)
• D. Does not exist

Hint:

Whenever logarithmic limits produce the indeterminate form \(\frac00\), L'Hospital's Rule is usually the fastest method: \[ \frac{d}{dx}(\log x)=\frac1x \]

Answer 1

The Correct Option is C

Concept: Whenever a limit takes the form: \[ \frac{0}{0} \] we may use either algebraic simplification or L'Hospital's Rule. L'Hospital's Rule states: \[ \lim_{x\to a}\frac{f(x)}{g(x)} = \lim_{x\to a}\frac{f'(x)}{g'(x)} \] provided both numerator and denominator approach zero.

Step 1: Check the indeterminate form.
Substituting \(x=e\): \[ \log e -1 =1-1=0 \] and \[ e-e=0 \] Hence the limit is of the form: \[ \frac00 \] Therefore, L'Hospital's Rule can be applied.

Step 2: Differentiate numerator and denominator.
Differentiate numerator: \[ \frac{d}{dx}(\log x-1)=\frac1x \] Differentiate denominator: \[ \frac{d}{dx}(x-e)=1 \] Thus, \[ \lim_{x\to e}\frac{\log x-1}{x-e} = \lim_{x\to e}\frac{1/x}{1} \] \[ = \lim_{x\to e}\frac1x \] \[ = \frac1e \] Hence, \[ \boxed{\frac1e} \]