`f(x)={{:((x-1)/(x+1)", "xne1),(lamda-1","x=1):}` at x = 1.

Answer

To determine the value of \( \lambda \) such that the function

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

is continuous at \( x = 1 \), we need to ensure that the left-hand limit, the right-hand limit, and the function value at \( x = 1 \) are all equal.

### Step 1: Find the left-hand limit as \( x \) approaches 1

The left-hand limit is given by:

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

Substituting \( x = 1 \):

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

### Step 2: Find the right-hand limit as \( x \) approaches 1

The right-hand limit is also given by:

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

Substituting \( x = 1 \):

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

### Step 3: Evaluate the function value at \( x = 1 \)

From the definition of the function:

\[ f(1) = \lambda - 1 \]

### Step 4: Set the limits equal to the function value for continuity

For \( f(x) \) to be continuous at \( x = 1 \), we require:

\[ \lim_{x \to 1} f(x) = f(1) \]

This gives us:

\[ 0 = \lambda - 1 \]

### Step 5: Solve for \( \lambda \)

From the equation \( 0 = \lambda - 1 \), we can solve for \( \lambda \):

\[ \lambda = 1 \]

### Conclusion

The value of \( \lambda \) that makes the function continuous at \( x = 1 \) is:

\[ \lambda = 1 \]

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