If `f(x)=sin|x|-e^(|x|)` then at `x=0,f(x)` is

Options

A. Continuous but not differentiable
B. Neither continuous nor differentiable
C. Both continuous and differentiable
D. Discontinuous but may be differentiable

Explanation

To solve the problem, we need to evaluate the function \( f(x) = \sin|x| - e^{|x|} \) at \( x = 0 \).

### Step-by-Step Solution:

1. **Evaluate \( f(0) \)**:
\[ f(0) = \sin|0| - e^{|0|} \]
Since \( |0| = 0 \):
\[ f(0) = \sin(0) - e^{0} \]
We know that \( \sin(0) = 0 \) and \( e^{0} = 1 \):
\[ f(0) = 0 - 1 = -1 \]

2. **Check Continuity at \( x = 0 \)**:
To check if \( f(x) \) is continuous at \( x = 0 \), we need to find the left-hand limit (LHL) and right-hand limit (RHL) as \( x \) approaches 0.

- **Right-Hand Limit (RHL)**:
\[ \text{RHL} = \lim_{h \to 0^+} f(h) = \lim_{h \to 0^+} \left( \sin|h| - e^{|h|} \right) \]
Since \( |h| = h \) for \( h > 0 \):
\[ \text{RHL} = \lim_{h \to 0^+} \left( \sin(h) - e^{h} \right) \]
Using the limits:
\[ \text{RHL} = \sin(0) - e^{0} = 0 - 1 = -1 \]

- **Left-Hand Limit (LHL)**:
\[ \text{LHL} = \lim_{h \to 0^-} f(h) = \lim_{h \to 0^-} \left( \sin|h| - e^{|h|} \right) \]
Since \( |h| = -h \) for \( h < 0 \):
\[ \text{LHL} = \lim_{h \to 0^-} \left( \sin(-h) - e^{-h} \right) \]
Using the properties of sine:
\[ \text{LHL} = \lim_{h \to 0^-} \left( -\sin(h) - e^{-h} \right) \]
Again, using the limits:
\[ \text{LHL} = -\sin(0) - e^{0} = 0 - 1 = -1 \]

Since both the RHL and LHL are equal to \( -1 \):
\[ \lim_{x \to 0} f(x) = -1 \]
Thus, \( f(0) = -1 \) and the limit as \( x \) approaches 0 is also \( -1 \). Therefore, \( f(x) \) is continuous at \( x = 0 \).

3. **Check Differentiability at \( x = 0 \)**:
We will check if the left-hand derivative (LHD) and right-hand derivative (RHD) at \( x = 0 \) are equal.

- **Right-Hand Derivative (RHD)**:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\left( \sin(h) - e^{h} \right) - (-1)}{h} \]
Simplifying:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{\sin(h) - e^{h} + 1}{h} \]
This can be split into two limits:
\[ \text{RHD} = \lim_{h \to 0^+} \frac{\sin(h)}{h} - \lim_{h \to 0^+} \frac{e^{h} - 1}{h} \]
Both limits evaluate to 1:
\[ \text{RHD} = 1 - 1 = 0 \]

- **Left-Hand Derivative (LHD)**:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{\left( \sin(-h) - e^{-h} \right) - (-1)}{h} \]
Simplifying:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{-\sin(h) - e^{-h} + 1}{h} \]
Similar to RHD:
\[ \text{LHD} = \lim_{h \to 0^-} \frac{-\sin(h)}{h} - \lim_{h \to 0^-} \frac{e^{-h} - 1}{h} \]
Both limits evaluate to -1:
\[ \text{LHD} = -1 - (-1) = 0 \]

Since RHD = LHD = 0, \( f(x) \) is differentiable at \( x = 0 \).

### Conclusion:
The value of \( f(0) \) is \( -1 \), and the function is both continuous and differentiable at \( x = 0 \).