If `f(x)={((sqrt(g(x))-1)/(sqrtx-1)"," , x ne 1),(1",",x=1):}` and `g'(1)=2, g(1)=1,` then `lim_(xrarr1)f(x)` is equal to

Options

A. 1
B. 3
C. 2
D. 4

Answer

To solve the problem, we need to find the limit of the function \( f(x) \) as \( x \) approaches 1. The function is defined as:

\[ f(x) = \begin{cases} \frac{\sqrt{g(x)} - 1}{\sqrt{x} - 1}, & x \neq 1 \\ 1, & x = 1 \end{cases} \]

We are given that \( g'(1) = 2 \) and \( g(1) = 1 \).

### Step-by-step Solution:

1. **Identify the limit to evaluate:**
We need to evaluate \( \lim_{x \to 1} f(x) \). Since \( f(x) \) is defined differently at \( x = 1 \), we will focus on the case when \( x \neq 1 \):
\[ \lim_{x \to 1} f(x) = \lim_{x \to 1} \frac{\sqrt{g(x)} - 1}{\sqrt{x} - 1} \]

2. **Substitute \( x = 1 \):**
If we directly substitute \( x = 1 \) into the limit, we get:
\[ \frac{\sqrt{g(1)} - 1}{\sqrt{1} - 1} = \frac{\sqrt{1} - 1}{1 - 1} = \frac{0}{0} \]
This is an indeterminate form \( \frac{0}{0} \), so we can apply L'Hôpital's Rule.

3. **Apply L'Hôpital's Rule:**
According to L'Hôpital's Rule, we differentiate the numerator and the denominator:
- The derivative of the numerator \( \sqrt{g(x)} - 1 \) is:
\[ \frac{1}{2\sqrt{g(x)}} \cdot g'(x) \]
- The derivative of the denominator \( \sqrt{x} - 1 \) is:
\[ \frac{1}{2\sqrt{x}} \]

Thus, we can rewrite the limit as:
\[ \lim_{x \to 1} \frac{\frac{1}{2\sqrt{g(x)}} g'(x)}{\frac{1}{2\sqrt{x}}} \]

4. **Simplify the limit:**
This simplifies to:
\[ \lim_{x \to 1} \frac{g'(x) \cdot \sqrt{x}}{\sqrt{g(x)}} \]

5. **Evaluate the limit at \( x = 1 \):**
Now we substitute \( x = 1 \):
\[ = \frac{g'(1) \cdot \sqrt{1}}{\sqrt{g(1)}} = \frac{g'(1) \cdot 1}{\sqrt{1}} = g'(1) = 2 \]

6. **Conclusion:**
Therefore, we find that:
\[ \lim_{x \to 1} f(x) = 2 \]

### Final Answer:
\[ \lim_{x \to 1} f(x) = 2 \]