`lim_(xrarr1^+)(sqrt(x^2-1)+sqrt(x-1))/(sqrt(x^2-1))=`

Options

A. `(1)/(2)`
B. `sqrt(2)+1`
C. 1
D. `1+(1)/(sqrt(2))`

Explanation

To solve the limit \( \lim_{x \to 1^+} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}} \), we will follow these steps:

### Step 1: Substitute the limit
First, we substitute \( x = 1 \) directly into the expression to see what we get:
\[ \sqrt{1^2 - 1} + \sqrt{1 - 1} = \sqrt{0} + \sqrt{0} = 0 \]
And for the denominator:
\[ \sqrt{1^2 - 1} = \sqrt{0} = 0 \]
This gives us the indeterminate form \( \frac{0}{0} \).

### Step 2: Simplify the expression
To resolve the indeterminate form, we can simplify the expression. We will divide the numerator and the denominator by \( \sqrt{x^2 - 1} \):
\[ \lim_{x \to 1^+} \frac{\sqrt{x^2 - 1}}{\sqrt{x^2 - 1}} + \frac{\sqrt{x - 1}}{\sqrt{x^2 - 1}} = \lim_{x \to 1^+} 1 + \frac{\sqrt{x - 1}}{\sqrt{x^2 - 1}} \]

### Step 3: Analyze the second term
Next, we need to analyze the limit of the second term \( \frac{\sqrt{x - 1}}{\sqrt{x^2 - 1}} \). We can rewrite \( \sqrt{x^2 - 1} \) as \( \sqrt{(x - 1)(x + 1)} \):
\[ \lim_{x \to 1^+} \frac{\sqrt{x - 1}}{\sqrt{(x - 1)(x + 1)}} = \lim_{x \to 1^+} \frac{\sqrt{x - 1}}{\sqrt{x - 1} \cdot \sqrt{x + 1}} = \lim_{x \to 1^+} \frac{1}{\sqrt{x + 1}} \]

### Step 4: Substitute \( x = 1 \)
Now we can substitute \( x = 1 \) into the simplified expression:
\[ \lim_{x \to 1^+} \frac{1}{\sqrt{1 + 1}} = \frac{1}{\sqrt{2}} = \frac{1}{\sqrt{2}} \]

### Step 5: Combine the results
Now we combine the results from Step 2 and Step 4:
\[ \lim_{x \to 1^+} \left( 1 + \frac{1}{\sqrt{2}} \right) = 1 + \frac{1}{\sqrt{2}} \]

### Final Answer
Thus, the final limit is:
\[ \lim_{x \to 1^+} \frac{\sqrt{x^2 - 1} + \sqrt{x - 1}}{\sqrt{x^2 - 1}} = 1 + \frac{1}{\sqrt{2}} \]