Questions: Evaluate lim as x approaches 0 of (sqrt(x+1))/(ln(x+1))

Solution Steps

First, substitute \( x = 0 \) into the expression to check if the limit can be directly evaluated: \[ \frac{\sqrt{0+1}}{\ln(0+1)} = \frac{1}{\ln(1)}. \] Since \( \ln(1) = 0 \), the expression becomes \( \frac{1}{0} \), which is undefined. This indicates that the limit cannot be directly evaluated and requires further analysis.

Since substituting \( x = 0 \) results in an indeterminate form \( \frac{1}{0} \), we can use L'Hôpital's Rule. L'Hôpital's Rule states that if: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \frac{0}{0} \text{ or } \frac{\infty}{\infty}, \] then: \[ \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)}, \] provided the limit on the right exists.

Here, let: \[ f(x) = \sqrt{x+1}, \quad g(x) = \ln(x+1). \] Compute the derivatives: \[ f'(x) = \frac{1}{2\sqrt{x+1}}, \quad g'(x) = \frac{1}{x+1}. \] Now, apply L'Hôpital's Rule: \[ \lim_{x \to 0} \frac{\sqrt{x+1}}{\ln(x+1)} = \lim_{x \to 0} \frac{f'(x)}{g'(x)} = \lim_{x \to 0} \frac{\frac{1}{2\sqrt{x+1}}}{\frac{1}{x+1}}. \]

Simplify the expression: \[ \lim_{x \to 0} \frac{\frac{1}{2\sqrt{x+1}}}{\frac{1}{x+1}} = \lim_{x \to 0} \frac{x+1}{2\sqrt{x+1}}. \] Factor out \( \sqrt{x+1} \) from the numerator: \[ \lim_{x \to 0} \frac{\sqrt{x+1} \cdot \sqrt{x+1}}{2\sqrt{x+1}} = \lim_{x \to 0} \frac{\sqrt{x+1}}{2}. \] Now, substitute \( x = 0 \): \[ \frac{\sqrt{0+1}}{2} = \frac{1}{2}. \]

Final Answer