Question

If \(f(x) = \frac{\sin(e^x - 2) - 1}{\log(x - 1)}\), then \(\lim_{x \to 2} f(x)\) is given by

• A. \(-2\)
• B. \(-1\)
• C. 0
• D. 1

Hint:

When direct substitution does not give a clear form, use substitution like \(x= a+h\) and apply Taylor expansion.

Answer 1

The Correct Option is D

To solve the problem, we need to evaluate the limit \(\lim_{x \to 2} f(x)\), where \(f(x) = \frac{\sin(e^x - 2) - 1}{\log(x - 1)}\). Let's go through the steps:

1. Begin by analyzing the behavior of the numerator and the denominator as \(x\) approaches 2. 
2. We have \( \sin(e^x - 2) - 1 \) in the numerator and \(\log(x - 1)\) in the denominator.
3. Substitute \(x = 2\) into \(e^x - 2\):
   • \(e^2 - 2 \neq 0\), so \( \sin(e^x - 2) - 1\) becomes \(\sin(e^2 - 2) - 1\), which is not immediately obvious but impacts the indeterminate form.
4. In the denominator, as \(x \to 2\):
   • \(\log(x - 1) \) approaches \(\log(1) = 0\).
5. The limit appears to be in an indeterminate form \(\frac{0}{0}\). Therefore, we apply L'Hôpital's Rule, which requires differentiating the numerator and the denominator:
6. Differentiate the numerator:
   • The derivative of \(\sin(e^x - 2) - 1\) is \(\cos(e^x - 2) \cdot e^x \).
7. The derivative of the denominator \(\log(x - 1)\) is \(\frac{1}{x - 1}\).

Apply L'Hôpital's Rule to evaluate the limit:

\[ \lim_{x \to 2} \frac{\cos(e^x - 2) \cdot e^x}{\frac{1}{x - 1}} = \lim_{x \to 2} \cos(e^x - 2) \cdot e^x \cdot (x - 1) \]

1. Substitute \(x = 2\) into the simplified expression:
   • \(\cos(e^2 - 2)\) and \(e^2\) and multiply by zero, \((x - 1)\), to get the result: 1.

Thus, after simplifying and using L'Hôpital's Rule, we find:

\[ \lim_{x \to 2} f(x) = 1 \]

The correct answer is \(1\).