Let `f(x) = (xe)^(1/|x|+1/x); x != 0, f(0) = 0`, test the continuity & differentiability at `x = 0`

Answer

To determine the continuity and differentiability of the function \( f(x) = x e^{\left(\frac{1}{|x|} + \frac{1}{x}\right)} \) for \( x \neq 0 \) and \( f(0) = 0 \) at \( x = 0 \), we will follow these steps:

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

To check the continuity of \( f(x) \) at \( x = 0 \), we need to find the left-hand limit (LHL) and the right-hand limit (RHL) as \( x \) approaches 0, and see if they equal \( f(0) \).

**Left-Hand Limit (LHL):**
\[ \text{LHL} = \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} x e^{\left(\frac{1}{|x|} + \frac{1}{x}\right)} = \lim_{x \to 0^-} x e^{\left(\frac{1}{-x} + \frac{1}{x}\right)} = \lim_{x \to 0^-} x e^{-\frac{1}{x} + \frac{1}{x}} = \lim_{x \to 0^-} x e^{0} = \lim_{x \to 0^-} x = 0 \]

**Right-Hand Limit (RHL):**
\[ \text{RHL} = \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} x e^{\left(\frac{1}{|x|} + \frac{1}{x}\right)} = \lim_{x \to 0^+} x e^{\left(\frac{1}{x} + \frac{1}{x}\right)} = \lim_{x \to 0^+} x e^{\frac{2}{x}} \]
As \( x \to 0^+ \), \( e^{\frac{2}{x}} \to \infty \), hence:
\[ \text{RHL} = \lim_{x \to 0^+} x e^{\frac{2}{x}} = 0 \]

Both LHL and RHL are equal to 0, and since \( f(0) = 0 \), we conclude that:
\[ \text{LHL} = \text{RHL} = f(0) \]
Thus, \( f(x) \) is continuous at \( x = 0 \).

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

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

**Left-Hand Derivative (LHD):**
\[ \text{LHD} = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{h e^{\left(\frac{1}{|h|} + \frac{1}{h}\right)} - 0}{h} = \lim_{h \to 0^-} e^{\left(\frac{1}{-h} + \frac{1}{h}\right)} = \lim_{h \to 0^-} e^{0} = 1 \]

**Right-Hand Derivative (RHD):**
\[ \text{RHD} = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h e^{\left(\frac{1}{|h|} + \frac{1}{h}\right)} - 0}{h} = \lim_{h \to 0^+} e^{\left(\frac{1}{h} + \frac{1}{h}\right)} = \lim_{h \to 0^+} e^{\frac{2}{h}} \]
As \( h \to 0^+ \), \( e^{\frac{2}{h}} \to \infty \), hence:
\[ \text{RHD} = \infty \]

Since LHD \( = 1 \) and RHD \( = \infty \), we find that:
\[ \text{LHD} \neq \text{RHD} \]
Thus, \( f(x) \) is not differentiable at \( x = 0 \).

### Conclusion
- The function \( f(x) \) is continuous at \( x = 0 \).
- The function \( f(x) \) is not differentiable at \( x = 0 \).