Evaluate the following limits :
`Lim_(x to 0) (sin^(2)5x)/(sin 15x)`

Answer

To evaluate the limit \( \lim_{x \to 0} \frac{\sin^2(5x)}{\sin(15x)} \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit expression:
\[ \lim_{x \to 0} \frac{\sin^2(5x)}{\sin(15x)} \]

### Step 2: Use the small angle approximation
As \( x \) approaches 0, we can use the fact that \( \sin(kx) \approx kx \) for small values of \( x \). Therefore, we can rewrite the sine functions:
\[ \sin(5x) \approx 5x \quad \text{and} \quad \sin(15x) \approx 15x \]

### Step 3: Substitute the approximations
Substituting these approximations into the limit gives:
\[ \lim_{x \to 0} \frac{(5x)^2}{15x} \]

### Step 4: Simplify the expression
Now simplify the expression:
\[ = \lim_{x \to 0} \frac{25x^2}{15x} = \lim_{x \to 0} \frac{25}{15} x = \lim_{x \to 0} \frac{5}{3} x \]

### Step 5: Evaluate the limit
As \( x \) approaches 0, the expression \( \frac{5}{3} x \) approaches:
\[ \frac{5}{3} \cdot 0 = 0 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{\sin^2(5x)}{\sin(15x)} = 0 \]

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