Discuss the continuity of f(x) at x = 0 if :
`f(x)={{:((sqrt(1+x)-sqrt(1-x))/(sinx)", if "xne0),(1" , if "x=0):}`.

Answer

To determine the continuity of the function \( f(x) \) at \( x = 0 \), we need to check three things:

1. The value of the function at \( x = 0 \).
2. The left-hand limit as \( x \) approaches \( 0 \).
3. The right-hand limit as \( x \) approaches \( 0 \).

If all three values are equal, then the function is continuous at that point.

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

Given:
\[ f(x) = \begin{cases} \frac{\sqrt{1+x} - \sqrt{1-x}}{\sin x} & \text{if } x \neq 0 \\ 1 & \text{if } x = 0 \end{cases} \]

At \( x = 0 \):
\[ f(0) = 1 \]

### Step 2: Find the Left-Hand Limit \( \lim_{x \to 0^-} f(x) \)

We need to evaluate:
\[ \lim_{x \to 0^-} f(x) = \lim_{x \to 0^-} \frac{\sqrt{1+x} - \sqrt{1-x}}{\sin x} \]

Substituting \( x = -h \) where \( h \to 0^+ \):
\[ \lim_{h \to 0} \frac{\sqrt{1 - h} - \sqrt{1 + h}}{-\sin h} = \lim_{h \to 0} \frac{\sqrt{1 - h} - \sqrt{1 + h}}{-\sin h} \]

Now, we can simplify the numerator by multiplying by the conjugate:
\[ \lim_{h \to 0} \frac{(\sqrt{1 - h} - \sqrt{1 + h})(\sqrt{1 - h} + \sqrt{1 + h})}{-\sin h (\sqrt{1 - h} + \sqrt{1 + h})} \]

This simplifies to:
\[ \lim_{h \to 0} \frac{(1 - h) - (1 + h)}{-\sin h (\sqrt{1 - h} + \sqrt{1 + h})} = \lim_{h \to 0} \frac{-2h}{-\sin h (\sqrt{1 - h} + \sqrt{1 + h})} \]

Now, we can use the limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \):
\[ = \lim_{h \to 0} \frac{2h}{\sin h (\sqrt{1 - h} + \sqrt{1 + h})} = \lim_{h \to 0} \frac{2}{\frac{\sin h}{h} (\sqrt{1 - h} + \sqrt{1 + h})} \]

As \( h \to 0 \), \( \sqrt{1 - h} \to 1 \) and \( \sqrt{1 + h} \to 1 \):
\[ = \frac{2}{1 \cdot (1 + 1)} = \frac{2}{2} = 1 \]

### Step 3: Find the Right-Hand Limit \( \lim_{x \to 0^+} f(x) \)

Now we evaluate:
\[ \lim_{x \to 0^+} f(x) = \lim_{x \to 0^+} \frac{\sqrt{1+x} - \sqrt{1-x}}{\sin x} \]

Substituting \( x = h \) where \( h \to 0^+ \):
\[ \lim_{h \to 0} \frac{\sqrt{1 + h} - \sqrt{1 - h}}{\sin h} \]

Using the same conjugate multiplication:
\[ = \lim_{h \to 0} \frac{(\sqrt{1 + h} - \sqrt{1 - h})(\sqrt{1 + h} + \sqrt{1 - h})}{\sin h (\sqrt{1 + h} + \sqrt{1 - h})} \]

This simplifies to:
\[ = \lim_{h \to 0} \frac{(1 + h) - (1 - h)}{\sin h (\sqrt{1 + h} + \sqrt{1 - h})} = \lim_{h \to 0} \frac{2h}{\sin h (\sqrt{1 + h} + \sqrt{1 - h})} \]

Using the limit \( \lim_{h \to 0} \frac{\sin h}{h} = 1 \):
\[ = \lim_{h \to 0} \frac{2}{\frac{\sin h}{h} (\sqrt{1 + h} + \sqrt{1 - h})} = \frac{2}{1 \cdot (1 + 1)} = 1 \]

### Conclusion

Now we have:
- \( f(0) = 1 \)
- \( \lim_{x \to 0^-} f(x) = 1 \)
- \( \lim_{x \to 0^+} f(x) = 1 \)

Since all three values are equal, we conclude that \( f(x) \) is continuous at \( x = 0 \).

### Final Answer
Thus, \( f(x) \) is continuous at \( x = 0 \).

---