`f(x) = {{:((sqrt(1+kx)-sqrt(1-kx))/(x),if -1 le x lt 0),((2x+1)/(x-1),if 0 le x le 1):}` at `x = 0`.

Explanation

To determine the value of \( k \) for which the function

\[ f(x) = \begin{cases} \frac{\sqrt{1+kx} - \sqrt{1-kx}}{x} & \text{if } -1 \leq x < 0 \\ \frac{2x + 1}{x - 1} & \text{if } 0 \leq x \leq 1 \end{cases} \]

is continuous at \( x = 0 \), we need to ensure that the left-hand limit as \( x \) approaches \( 0 \) from the left equals the value of the function at \( x = 0 \), which in turn equals the right-hand limit as \( x \) approaches \( 0 \) from the right.

### Step 1: Find \( f(0) \)

For \( x = 0 \), we use the second case of the piecewise function:

\[ f(0) = \frac{2(0) + 1}{0 - 1} = \frac{1}{-1} = -1 \]

### Step 2: Find the left-hand limit as \( x \to 0^- \)

We calculate the limit of the first case as \( x \) approaches \( 0 \) from the left:

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

To simplify this limit, we can multiply the numerator and denominator by the conjugate:

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

This simplifies to:

\[ \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2k}{\sqrt{1+kx} + \sqrt{1-kx}} \]

### Step 3: Evaluate the limit as \( x \to 0 \)

As \( x \) approaches \( 0 \):

\[ \sqrt{1+kx} \to \sqrt{1} = 1 \quad \text{and} \quad \sqrt{1-kx} \to \sqrt{1} = 1 \]

Thus, we have:

\[ \lim_{x \to 0^-} \frac{2k}{\sqrt{1+kx} + \sqrt{1-kx}} = \frac{2k}{1 + 1} = \frac{2k}{2} = k \]

### Step 4: Find the right-hand limit as \( x \to 0^+ \)

Now we calculate the limit from the right:

\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{2x + 1}{x - 1} = \frac{2(0) + 1}{0 - 1} = \frac{1}{-1} = -1 \]

### Step 5: Set the limits equal for continuity

For \( f(x) \) to be continuous at \( x = 0 \), we set the left-hand limit equal to \( f(0) \):

\[ k = -1 \]

### Conclusion

Thus, the value of \( k \) for which \( f(x) \) is continuous at \( x = 0 \) is:

\[ \boxed{-1} \]