Prove that the function If `f(x)={:{((x)/(1+e^(1//x)) ", " x ne 0),(" "0", " x=0):}` is not differentiable

Options

A. Differentiable
B. Non-differentiable
C. L.H.L at x-0 is 1
D. None of these

Answer

To prove that the function

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

is not differentiable at \( x = 0 \), we will calculate the left-hand derivative (LHD) and the right-hand derivative (RHD) at \( x = 0 \) and show that they are not equal.

### Step 1: Calculate the Left-Hand Derivative (LHD)

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

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

Since \( f(0) = 0 \), we can substitute this into the equation:

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

Now, for \( x = -h \) (where \( h > 0 \)), we have:

\[ f(-h) = \frac{-h}{1 + e^{-\frac{1}{h}}} \]

Substituting this into the LHD expression gives:

\[ LHD = \lim_{h \to 0^-} \frac{\frac{-h}{1 + e^{-\frac{1}{h}}}}{h} \]

This simplifies to:

\[ LHD = \lim_{h \to 0^-} \frac{-1}{1 + e^{-\frac{1}{h}}} \]

As \( h \to 0^- \), \( e^{-\frac{1}{h}} \to 0 \) (since \(-\frac{1}{h} \to -\infty\)), so we have:

\[ LHD = \frac{-1}{1 + 0} = -1 \]

### Step 2: Calculate the Right-Hand Derivative (RHD)

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

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

Again, since \( f(0) = 0 \), we can substitute this into the equation:

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

For \( x = h \) (where \( h > 0 \)), we have:

\[ f(h) = \frac{h}{1 + e^{\frac{1}{h}}} \]

Substituting this into the RHD expression gives:

\[ RHD = \lim_{h \to 0^+} \frac{\frac{h}{1 + e^{\frac{1}{h}}}}{h} \]

This simplifies to:

\[ RHD = \lim_{h \to 0^+} \frac{1}{1 + e^{\frac{1}{h}}} \]

As \( h \to 0^+ \), \( e^{\frac{1}{h}} \to \infty \) (since \(\frac{1}{h} \to \infty\)), so we have:

\[ RHD = \frac{1}{1 + \infty} = 0 \]

### Step 3: Conclusion

Now we have:

\[ LHD = -1 \quad \text{and} \quad RHD = 0 \]

Since \( LHD \neq RHD \), we conclude that the function \( f(x) \) is not differentiable at \( x = 0 \).