Which of the following is true about
`Let f(x) ={{:((x-2)/(|x-2|)((x^(2)-1)/(x^(2)+1)),xne2),((3)/(5),x=2):}?`

Options

A. f(X) is continuous at x=2
B. f(x) has removable discontinuity ata x=2.
C. f(x) has non-removable discontinuity at x=2 .
D. Discontiuity at x=2 can be removed by redefining the function at x=2.

Answer

To determine the continuity of the function

\[ f(x) = \begin{cases} \frac{(x-2)}{|x-2|} \cdot \frac{(x^2-1)}{(x^2+1)}, & x \neq 2 \\ \frac{3}{5}, & x = 2 \end{cases} \]

at \( x = 2 \), we need to check the following condition for continuity:

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

### Step 1: Calculate \( f(2) \)

From the definition, when \( x = 2 \):

\[ f(2) = \frac{3}{5} \]

### Step 2: Calculate \( \lim_{x \to 2^+} f(x) \)

For \( x \to 2^+ \) (approaching 2 from the right), we have \( |x-2| = x-2 \). Therefore,

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

Now, we compute the limit:

\[ \lim_{x \to 2^+} f(x) = \lim_{x \to 2} \frac{x^2-1}{x^2+1} = \frac{2^2-1}{2^2+1} = \frac{4-1}{4+1} = \frac{3}{5} \]

### Step 3: Calculate \( \lim_{x \to 2^-} f(x) \)

For \( x \to 2^- \) (approaching 2 from the left), we have \( |x-2| = -(x-2) = 2-x \). Therefore,

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

Now, we compute the limit:

\[ \lim_{x \to 2^-} f(x) = \lim_{x \to 2} -\frac{x^2-1}{x^2+1} = -\frac{2^2-1}{2^2+1} = -\frac{4-1}{4+1} = -\frac{3}{5} \]

### Step 4: Compare the limits and \( f(2) \)

Now we compare the limits and the function value at \( x = 2 \):

- \( \lim_{x \to 2^+} f(x) = \frac{3}{5} \)
- \( \lim_{x \to 2^-} f(x) = -\frac{3}{5} \)
- \( f(2) = \frac{3}{5} \)

Since \( \lim_{x \to 2^-} f(x) \neq \lim_{x \to 2^+} f(x) \), the function is not continuous at \( x = 2 \).

### Conclusion

Thus, the function \( f(x) \) is discontinuous at \( x = 2 \).