If `f(x) ={:{((sin 3x)/(sin 5x)", "x ne 0),(0", " x= 0):},` then discuss its continuity at x= 0. .

Answer

To discuss the continuity of the function

\[ f(x) = \begin{cases} \frac{\sin(3x)}{\sin(5x)} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases} \]

at \( x = 0 \), we need to check the following condition for continuity:

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

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

From the definition of the function, we have:

\[ f(0) = 0 \]

### Step 2: Calculate \( \lim_{x \to 0} f(x) \)

For \( x \neq 0 \), we need to evaluate:

\[ \lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)} \]

### Step 3: Apply the limit

Since both the numerator and denominator approach 0 as \( x \) approaches 0, we have an indeterminate form \( \frac{0}{0} \). We can use the following trigonometric limit identity:

\[ \lim_{x \to 0} \frac{\sin(kx)}{kx} = 1 \]

We can rewrite the limit as:

\[ \lim_{x \to 0} \frac{\sin(3x)}{\sin(5x)} = \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{3x}{5x} \cdot \frac{5x}{\sin(5x)} \]

### Step 4: Simplify the expression

This can be rearranged as:

\[ \lim_{x \to 0} \frac{\sin(3x)}{3x} \cdot \frac{3}{5} \cdot \lim_{x \to 0} \frac{5x}{\sin(5x)} \]

### Step 5: Evaluate the limits

Using the limit identity mentioned earlier:

\[ \lim_{x \to 0} \frac{\sin(3x)}{3x} = 1 \]
\[ \lim_{x \to 0} \frac{5x}{\sin(5x)} = 1 \]

Thus, we have:

\[ \lim_{x \to 0} f(x) = 1 \cdot \frac{3}{5} \cdot 1 = \frac{3}{5} \]

### Step 6: Compare the limits

Now we compare the limit with the value of the function at \( x = 0 \):

\[ \lim_{x \to 0} f(x) = \frac{3}{5} \quad \text{and} \quad f(0) = 0 \]

### Conclusion

Since

\[ \lim_{x \to 0} f(x) \neq f(0) \]

the function \( f(x) \) is discontinuous at \( x = 0 \).

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