If `f(x) = x [sqrt(x) - sqrt(x+1)]`, then

Options

A. f(x) is continuous but not differentiable at x = 0
B. f(x) is continuous and differentiable at x=0
C. f(x) is not differentiable at x = 0
D. none of these

Answer

To solve the problem of determining the continuity and differentiability of the function \( f(x) = x [\sqrt{x} - \sqrt{x+1}] \) at \( x = 0 \), we will follow these steps:

### Step 1: Evaluate the function at \( x = 0 \)
First, we need to find the value of the function at \( x = 0 \):
\[ f(0) = 0 \left[\sqrt{0} - \sqrt{0 + 1}\right] = 0 \left[0 - 1\right] = 0 \]

### Step 2: Find the left-hand limit as \( x \) approaches 0
Next, we will find the left-hand limit of \( f(x) \) as \( x \) approaches 0. Since we are interested in the left-hand limit, we will consider values of \( x \) that are slightly less than 0 (i.e., \( x \to 0^- \)):
\[ \lim_{h \to 0^+} f(0 - h) = \lim_{h \to 0^+} f(-h) \]
Substituting \( -h \) into the function:
\[ f(-h) = -h \left[\sqrt{-h} - \sqrt{-h + 1}\right] \]
However, since \( \sqrt{-h} \) is not defined for \( h > 0 \) (as it results in a complex number), we conclude that:
\[ \lim_{h \to 0^+} f(-h) \text{ does not exist.} \]

### Step 3: Find the right-hand limit as \( x \) approaches 0
Now, we will find the right-hand limit as \( x \) approaches 0:
\[ \lim_{h \to 0^+} f(h) = \lim_{h \to 0^+} h \left[\sqrt{h} - \sqrt{h + 1}\right] \]
Calculating this limit:
\[ = \lim_{h \to 0^+} h \left[\sqrt{h} - \sqrt{1}\right] = \lim_{h \to 0^+} h \left[\sqrt{h} - 1\right] \]
As \( h \to 0^+ \), \( \sqrt{h} \to 0 \):
\[ = \lim_{h \to 0^+} h \left[0 - 1\right] = \lim_{h \to 0^+} -h = 0 \]

### Step 4: Check continuity at \( x = 0 \)
For \( f(x) \) to be continuous at \( x = 0 \), the following must hold:
\[ \lim_{x \to 0} f(x) = f(0) \]
We found that:
- \( f(0) = 0 \)
- The left-hand limit does not exist.
- The right-hand limit is \( 0 \).

Since the left-hand limit does not exist, we conclude that:
\[ f(x) \text{ is not continuous at } x = 0. \]

### Step 5: Check differentiability at \( x = 0 \)
A function must be continuous at a point to be differentiable there. Since \( f(x) \) is not continuous at \( x = 0 \), it cannot be differentiable at that point.

### Conclusion
Thus, we conclude that:
- The function \( f(x) \) is not continuous at \( x = 0 \).
- The function \( f(x) \) is not differentiable at \( x = 0 \).