If `f(x) ={{:(x(e)^(-[1/|x|+1/x]), x ne 0),(0, x=0):}`, then f(x) is:

Options

A. continuous and differentiable `AA` x
B. continuous for `AA` x but not differentiable at x = 0
C. neither differentiable nor continuous at x = 0
D. discontinuous everywhere

Answer

To determine the continuity and differentiability of the function \( f(x) \) defined as:

\[ f(x) = \begin{cases} x e^{-\left(\frac{1}{|x|} + \frac{1}{x}\right)}, & x \neq 0 \\ 0, & x = 0 \end{cases} \]

we will analyze the function at the point \( x = 0 \).

### Step 1: Check Continuity at \( x = 0 \)

To check if \( f(x) \) is continuous at \( x = 0 \), we need to verify that:

\[ \lim_{x \to 0} f(x) = f(0) \]

Since \( f(0) = 0 \), we need to find the left-hand limit and the right-hand limit as \( x \) approaches 0.

#### Left-Hand Limit:

For \( x < 0 \):

\[ f(x) = x e^{-\left(\frac{1}{-x} + \frac{1}{x}\right)} = x e^{-\left(-\frac{1}{x} + \frac{1}{x}\right)} = x e^{0} = x \]

Now we calculate the limit:

\[ \lim_{h \to 0^-} f(h) = \lim_{h \to 0^-} h = 0 \]

#### Right-Hand Limit:

For \( x > 0 \):

\[ f(x) = x e^{-\left(\frac{1}{x} + \frac{1}{x}\right)} = x e^{-\frac{2}{x}} \]

Now we calculate the limit:

\[ \lim_{h \to 0^+} f(h) = \lim_{h \to 0^+} h e^{-\frac{2}{h}} \]

As \( h \to 0^+ \), \( e^{-\frac{2}{h}} \to 0 \) much faster than \( h \to 0 \), so:

\[ \lim_{h \to 0^+} h e^{-\frac{2}{h}} = 0 \]

### Conclusion on Continuity:

Since both the left-hand limit and right-hand limit as \( x \to 0 \) are equal to \( 0 \), we have:

\[ \lim_{x \to 0} f(x) = 0 = f(0) \]

Thus, \( f(x) \) is continuous at \( x = 0 \).

### Step 2: Check Differentiability at \( x = 0 \)

To check differentiability at \( x = 0 \), we need to find the left-hand derivative and the right-hand derivative.

#### Left-Hand Derivative:

The left-hand derivative at \( x = 0 \) is given by:

\[ f'_{-}(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h - 0} = \lim_{h \to 0^-} \frac{h - 0}{h} = \lim_{h \to 0^-} 1 = 1 \]

#### Right-Hand Derivative:

The right-hand derivative at \( x = 0 \) is given by:

\[ f'_{+}(0) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h - 0} = \lim_{h \to 0^+} \frac{h e^{-\frac{2}{h}} - 0}{h} = \lim_{h \to 0^+} e^{-\frac{2}{h}} \]

As \( h \to 0^+ \), \( e^{-\frac{2}{h}} \to 0 \), so:

\[ f'_{+}(0) = 0 \]

### Conclusion on Differentiability:

Since the left-hand derivative \( f'_{-}(0) = 1 \) and the right-hand derivative \( f'_{+}(0) = 0 \) are not equal, \( f(x) \) is not differentiable at \( x = 0 \).

### Final Result:

- \( f(x) \) is continuous for all \( x \).
- \( f(x) \) is not differentiable at \( x = 0 \).