The value of f(0) so that the function `f(x)=(1)/(x)-(1)/(sinx)` is continuous at x = 0 is ________

Answer

To find the value of \( f(0) \) so that the function

\[ f(x) = \frac{1}{x} - \frac{1}{\sin x} \]

is continuous at \( x = 0 \), we need to evaluate the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Find the limit of \( f(x) \) as \( x \) approaches 0

We start by rewriting \( f(x) \):

\[ f(x) = \frac{1}{x} - \frac{1}{\sin x} \]

To find the limit as \( x \) approaches 0, we can combine the two fractions:

\[ f(x) = \frac{\sin x - x}{x \sin x} \]

### Step 2: Evaluate the limit

Now we need to evaluate:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x - x}{x \sin x} \]

As \( x \) approaches 0, both the numerator and denominator approach 0, which gives us a \( \frac{0}{0} \) indeterminate form. We can apply L'Hôpital's Rule.

### Step 3: Apply L'Hôpital's Rule

Differentiating the numerator and the denominator:

- The derivative of the numerator \( \sin x - x \) is \( \cos x - 1 \).
- The derivative of the denominator \( x \sin x \) is \( \sin x + x \cos x \).

Thus, we have:

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

### Step 4: Evaluate the new limit

Substituting \( x = 0 \):

- The numerator becomes \( \cos(0) - 1 = 1 - 1 = 0 \).
- The denominator becomes \( \sin(0) + 0 \cdot \cos(0) = 0 + 0 = 0 \).

This is again a \( \frac{0}{0} \) form, so we apply L'Hôpital's Rule again.

### Step 5: Differentiate again

Differentiating again:

- The derivative of the numerator \( \cos x - 1 \) is \( -\sin x \).
- The derivative of the denominator \( \sin x + x \cos x \) is \( \cos x + \cos x - x \sin x = 2\cos x - x \sin x \).

Now we have:

\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{-\sin x}{2\cos x - x \sin x} \]

### Step 6: Substitute \( x = 0 \)

Substituting \( x = 0 \):

- The numerator becomes \( -\sin(0) = 0 \).
- The denominator becomes \( 2\cos(0) - 0 \cdot \sin(0) = 2 \cdot 1 - 0 = 2 \).

Thus, we find:

\[ \lim_{x \to 0} f(x) = \frac{0}{2} = 0 \]

### Conclusion

For the function \( f(x) \) to be continuous at \( x = 0 \), we must have:

\[ f(0) = 0 \]

Therefore, the value of \( f(0) \) so that the function is continuous at \( x = 0 \) is:

\[ \boxed{0} \]