Question

Evaluate the limit: $\lim_{x \to 1} \left[ \frac{\sqrt{x} - 1}{\log x} \right] =$

• A. $\frac{1}{2}$
• B. 2
• C. $-2$
• D. $-\frac{1}{2}$

Hint:

You can also solve this using algebraic substitution! Let $x = t^2$. As $x \to 1$, $t \to 1$. The limit transforms into $\lim_{t \to 1} \frac{t-1}{\log(t^2)} = \lim_{t \to 1} \frac{t-1}{2\log t}$. Since $\lim_{t \to 1} \frac{\log t}{t-1} = 1$ is a standard fundamental limit property, its reciprocal is also 1, leaving you with just the constant factor $\frac{1}{2}$ in front.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
The problem requires evaluating the limit of a logarithmic fraction as $x$ approaches 1.

Step 2: Key Formula or Approach:
Substituting $x = 1$ directly into the expression yields: $$\frac{\sqrt{1} - 1}{\log(1)} = \frac{0}{0}$$ This is a standard indeterminate form, which allows us to apply L'Hôpital's Rule. This rule states that the limit of a fraction of functions is equal to the limit of the fraction of their derivatives: $$\lim_{x \to c} \frac{f(x)}{g(x)} = \lim_{x \to c} \frac{f'(x)}{g'(x)}$$

Step 3: Detailed Explanation:
1. Find the derivative of the numerator function $f(x) = \sqrt{x} - 1$: $$f'(x) = \frac{1}{2\sqrt{x}}$$ 2. Find the derivative of the denominator function $g(x) = \log x$: $$g'(x) = \frac{1}{x}$$ 3. Apply L'Hôpital's Rule by substituting these derivatives back into the limit expression: $$\lim_{x \to 1} \frac{\sqrt{x} - 1}{\log x} = \lim_{x \to 1} \frac{\frac{1}{2\sqrt{x}}}{\frac{1}{x}}$$ 4. Simplify the complex fraction: $$= \lim_{x \to 1} \frac{x}{2\sqrt{x}} = \lim_{x \to 1} \frac{\sqrt{x}}{2}$$ 5. Substitute $x = 1$ into this simplified expression to evaluate the final limit: $$= \frac{\sqrt{1}}{2} = \frac{1}{2}$$

Step 4: Final Answer:
The evaluated limit value is $\frac{1}{2}$, which matches option (A).