Show the function, `f(x) = {{:((e^(1//x)-1)/(e^(1//x)+1)",","when",x ne 0),(0",","when",x = 0):}` has non-removable discontinuity at x = 0

Answer

To show that the function

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

has a non-removable discontinuity at \( x = 0 \), we will evaluate the left-hand limit and the right-hand limit as \( x \) approaches 0.

### Step 1: Calculate the Left-Hand Limit

We need to find

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

For \( x \) approaching 0 from the left (negative values), we have:

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

As \( x \to 0^- \), \( \frac{1}{x} \to -\infty \). Therefore,

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

Substituting this into our function gives:

\[ f(x) = \frac{0 - 1}{0 + 1} = \frac{-1}{1} = -1. \]

Thus,

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

### Step 2: Calculate the Right-Hand Limit

Next, we find

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

For \( x \) approaching 0 from the right (positive values), we again have:

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

As \( x \to 0^+ \), \( \frac{1}{x} \to +\infty \). Therefore,

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

Substituting this into our function gives:

\[ f(x) = \frac{\infty - 1}{\infty + 1} = \frac{\infty}{\infty}. \]

This is an indeterminate form, so we can apply L'Hôpital's Rule. We differentiate the numerator and denominator:

1. The derivative of the numerator \( e^{\frac{1}{x}} - 1 \) is \( e^{\frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right) \).
2. The derivative of the denominator \( e^{\frac{1}{x}} + 1 \) is also \( e^{\frac{1}{x}} \cdot \left(-\frac{1}{x^2}\right) \).

Applying L'Hôpital's Rule:

\[ \lim_{x \to 0^+} \frac{e^{\frac{1}{x}} - 1}{e^{\frac{1}{x}} + 1} = \lim_{x \to 0^+} \frac{-\frac{e^{\frac{1}{x}}}{x^2}}{-\frac{e^{\frac{1}{x}}}{x^2}} = \lim_{x \to 0^+} 1 = 1. \]

Thus,

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

### Step 3: Conclusion

Now we have:

\[ \lim_{x \to 0^-} f(x) = -1 \quad \text{and} \quad \lim_{x \to 0^+} f(x) = 1. \]

Since the left-hand limit and the right-hand limit are not equal:

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

we conclude that the function \( f(x) \) has a non-removable discontinuity at \( x = 0 \).