Let `f (x)= [{:((x(3e^(1//x)+4))/(2 -e^(1//x)),,,( x ne 0)),(0 ,,, x=0):} x ne (1)/(ln 2)` which of the following statement (s) is/are correct ?

Options

A. f (x) is continous `at x=0`
B. ` f (x)` is non-dervable at `x=0`
C. `f'(0^(+))=-3`
D. `f '(0^(-))` does not exist

Answer

To solve the problem, we need to analyze the function given:

\[ f(x) = \begin{cases} \frac{x(3e^{1/x} + 4)}{2 - e^{1/x}} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

We need to check the continuity and differentiability of \( f(x) \) at \( x = 0 \).

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

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

#### Left-Hand Limit:

\[ \lim_{h \to 0^-} f(h) = \lim_{h \to 0^-} \frac{h(3e^{1/h} + 4)}{2 - e^{1/h}} \]

As \( h \to 0^- \), \( e^{1/h} \to 0 \). Thus, we have:

\[ = \lim_{h \to 0^-} \frac{h(3 \cdot 0 + 4)}{2 - 0} = \lim_{h \to 0^-} \frac{4h}{2} = \lim_{h \to 0^-} 2h = 0 \]

#### Right-Hand Limit:

\[ \lim_{h \to 0^+} f(h) = \lim_{h \to 0^+} \frac{h(3e^{1/h} + 4)}{2 - e^{1/h}} \]

As \( h \to 0^+ \), \( e^{1/h} \to \infty \). Thus, we have:

\[ = \lim_{h \to 0^+} \frac{h(3 \cdot \infty + 4)}{2 - \infty} \approx \lim_{h \to 0^+} \frac{3h \cdot \infty}{-\infty} = 0 \]

Both limits equal 0, which is \( f(0) \). Therefore, \( f(x) \) 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:

\[ f'(0^-) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{\frac{h(3e^{1/h} + 4)}{2 - e^{1/h}} - 0}{h} \]

This simplifies to:

\[ = \lim_{h \to 0^-} \frac{3e^{1/h} + 4}{2 - e^{1/h}} = \frac{4}{2} = 2 \]

#### Right-Hand Derivative:

\[ f'(0^+) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{\frac{h(3e^{1/h} + 4)}{2 - e^{1/h}} - 0}{h} \]

This simplifies to:

\[ = \lim_{h \to 0^+} \frac{3e^{1/h} + 4}{2 - e^{1/h}} = \lim_{h \to 0^+} \frac{3 \cdot \infty + 4}{2 - \infty} = -\infty \]

### Conclusion

Since the left-hand derivative \( f'(0^-) = 2 \) and the right-hand derivative \( f'(0^+) = -\infty \) are not equal, \( f(x) \) is not differentiable at \( x = 0 \).

### Final Answer

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