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

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\sin^2(5x)}{\sin^2(bx)}, \]

we will follow these steps:

### Step 1: Check for Indeterminate Form
First, we substitute \(x = 0\) into the expression:

\[ \sin^2(5 \cdot 0) = \sin^2(0) = 0, \]
\[ \sin^2(b \cdot 0) = \sin^2(0) = 0. \]

Since both the numerator and denominator approach 0, we have an indeterminate form of type \(\frac{0}{0}\).

**Hint:** Always check if substituting the limit point gives an indeterminate form before proceeding with further calculations.

### Step 2: Use the Standard Limit
We will use the standard limit:

\[ \lim_{x \to 0} \frac{\sin x}{x} = 1. \]

To apply this, we rewrite the limit:

\[ \lim_{x \to 0} \frac{\sin^2(5x)}{\sin^2(bx)} = \lim_{x \to 0} \left( \frac{\sin(5x)}{5x} \cdot \frac{5x}{bx} \cdot \frac{bx}{\sin(bx)} \right)^2. \]

### Step 3: Rewrite the Expression
We rewrite the limit as follows:

\[ \lim_{x \to 0} \frac{\sin^2(5x)}{(5x)^2} \cdot \frac{(5x)^2}{(bx)^2} \cdot \frac{(bx)^2}{\sin^2(bx)}. \]

This can be expressed as:

\[ \lim_{x \to 0} \left( \frac{\sin(5x)}{5x} \right)^2 \cdot \frac{25x^2}{b^2x^2} \cdot \left( \frac{bx}{\sin(bx)} \right)^2. \]

### Step 4: Evaluate Each Limit
Now we evaluate each part separately:

1. \(\lim_{x \to 0} \frac{\sin(5x)}{5x} = 1\), so \(\left( \frac{\sin(5x)}{5x} \right)^2 \to 1^2 = 1\).
2. The term \(\frac{25x^2}{b^2x^2} = \frac{25}{b^2}\).
3. \(\lim_{x \to 0} \frac{bx}{\sin(bx)} = 1\), so \(\left( \frac{bx}{\sin(bx)} \right)^2 \to 1^2 = 1\).

### Step 5: Combine the Results
Combining all these results, we have:

\[ \lim_{x \to 0} \frac{\sin^2(5x)}{\sin^2(bx)} = 1 \cdot \frac{25}{b^2} \cdot 1 = \frac{25}{b^2}. \]

### Final Answer
Thus, the limit evaluates to:

\[ \boxed{\frac{25}{b^2}}. \]