Question

Evaluate the following limit:
\[ \lim_{x \to 0} \frac{\ln(1+x)}{2 \sin(x)} \quad \text{(rounded off to two decimal places).} \]

Hint:

L'Hopital's Rule is helpful for resolving indeterminate forms like \( \frac{0}{0} \) or \( \frac{\infty}{\infty} \) by differentiating the numerator and denominator.

Answer 1

Step 1: Identifying the limit form.
We need to evaluate the limit: \[ \lim_{x \to 0} \frac{\ln(1+x)}{2 \sin(x)}. \] At \(x = 0\), both the numerator and the denominator approach 0. Hence, we have a \( \frac{0}{0} \) indeterminate form, which suggests that we can apply L'Hopital's Rule.

Step 2: Applying L'Hopital's Rule.
L'Hopital's Rule states that if the limit is of the form \( \frac{0}{0} \), we can differentiate the numerator and denominator separately and then evaluate the limit. The derivative of the numerator \( \ln(1+x) \) is: \[ \frac{d}{dx} \ln(1+x) = \frac{1}{1+x}. \] The derivative of the denominator \( 2 \sin(x) \) is: \[ \frac{d}{dx} 2 \sin(x) = 2 \cos(x). \] Now, applying L'Hopital's Rule: \[ \lim_{x \to 0} \frac{\ln(1+x)}{2 \sin(x)} = \lim_{x \to 0} \frac{\frac{1}{1+x}}{2 \cos(x)}. \]

Step 3: Evaluating the new limit.
At \( x = 0 \), the numerator becomes \( \frac{1}{1+0} = 1 \), and the denominator becomes \( 2 \cos(0) = 2 \). Therefore, the limit simplifies to: \[ \frac{1}{2} = 0.50. \]

Thus, the value of the limit is: \[ \boxed{0.50}. \]