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

Options

A. (a)Continuous for all x but not differentiable
B. (b)Neither differentiable nor continuous
C. (c)Discontinuous everywhere
D. (d)Continuous as well as differentiable for all x

Answer

To determine the properties 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 need to check if the function is continuous and differentiable at \( x = 0 \).

### 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 compute \( \lim_{x \to 0} f(x) \).

For \( x \neq 0 \):

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

### Step 2: Evaluate the Limit as \( x \to 0 \)

We can analyze the limit from both sides (as \( x \to 0^+ \) and \( x \to 0^- \)).

**For \( x \to 0^+ \):**

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

As \( x \to 0^+ \), \( e^{-\frac{2}{x}} \) approaches \( 0 \) much faster than \( x \) approaches \( 0 \). Therefore:

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

**For \( x \to 0^- \):**

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

As \( x \to 0^- \), \( e^{\frac{2}{x}} \) approaches \( \infty \), but since \( x \) is negative, \( f(x) \) approaches \( 0 \):

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

### Step 3: Conclusion on Continuity

Since both one-sided limits equal \( 0 \):

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

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

### Step 4: Check Differentiability at \( x = 0 \)

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

**Left-hand derivative:**

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

**Right-hand derivative:**

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

### Step 5: Conclusion on Differentiability

Since the left-hand derivative does not equal the right-hand derivative:

\[ f'(0^-) \neq f'(0^+) \]

Thus, \( f(x) \) is not differentiable at \( x = 0 \).

### Final Conclusion

The function \( f(x) \) is continuous everywhere but not differentiable at \( x = 0 \).

### Answer

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