If `f(x)={:{((e^(1/x)-1)/(e^(1/x)+1)", for " x !=0),(1", for " x=0):}`, then f is

Options

A. continuous at `x=0`
B. discontinuous at `x=0`
C. continuous if `f(0)=-1`
D. discontinuous if `f(0)=-1`

Answer

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

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

we need to check the left-hand limit, right-hand limit, and the value of the function at \( x = 0 \).

### Step 1: Calculate the left-hand limit as \( x \) approaches 0 from the negative side.

We need to find:

\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{e^{1/x} - 1}{e^{1/x} + 1} \]

As \( x \) approaches 0 from the left (negative side), \( \frac{1}{x} \) approaches \( -\infty \). Therefore, we can evaluate:

\[ e^{1/x} \to e^{-\infty} = 0 \]

Substituting this into our limit:

\[ \lim_{x \to 0^-} f(x) = \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1 \]

### Step 2: Calculate the right-hand limit as \( x \) approaches 0 from the positive side.

Next, we find:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{e^{1/x} - 1}{e^{1/x} + 1} \]

As \( x \) approaches 0 from the right (positive side), \( \frac{1}{x} \) approaches \( +\infty \). Thus, we have:

\[ e^{1/x} \to e^{+\infty} = \infty \]

Substituting this into our limit:

\[ \lim_{x \to 0^+} f(x) = \frac{\infty - 1}{\infty + 1} = \frac{\infty}{\infty} = 1 \]

### Step 3: Evaluate the value of the function at \( x = 0 \).

From the definition of the function, we have:

\[ f(0) = 1 \]

### Step 4: Check for continuity at \( x = 0 \).

For the function \( f(x) \) to be continuous at \( x = 0 \), the following must hold:

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

From our calculations:

- \( \lim_{x \to 0^-} f(x) = -1 \)
- \( \lim_{x \to 0^+} f(x) = 1 \)
- \( f(0) = 1 \)

Since the left-hand limit does not equal the right-hand limit, and the left-hand limit does not equal \( f(0) \), we conclude that:

\[ f(x) \text{ is discontinuous at } x = 0. \]

### Final Conclusion:

Thus, the function \( f(x) \) is discontinuous at \( x = 0 \).

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