What is `lim_(xto0) (sin^(2)ax)/(bx)` (a,b are constants) equal to ?

Options

A. 0
B. a
C. a/b
D. Does not exist

Explanation

To solve the limit \( \lim_{x \to 0} \frac{\sin^2(ax)}{bx} \), where \( a \) and \( b \) are constants, we can follow these steps:

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

### Step 2: Factor out constants
We can factor out the constant \( b \) from the denominator:
\[ = \frac{1}{b} \lim_{x \to 0} \frac{\sin^2(ax)}{x} \]

### Step 3: Use the limit property of sine
We know from the standard limit that \( \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \). To use this, we need to manipulate our expression to match this form. We can rewrite \( \sin^2(ax) \) as:
\[ \sin^2(ax) = \left(\sin(ax)\right)^2 \]
Thus, we can express the limit as:
\[ = \frac{1}{b} \lim_{x \to 0} \frac{\sin(ax)}{x} \cdot \sin(ax) \]

### Step 4: Change of variable
To apply the limit property, we can change the variable. Let \( u = ax \). Then, as \( x \to 0 \), \( u \to 0 \) as well. We can express \( x \) in terms of \( u \):
\[ x = \frac{u}{a} \]
Substituting this into our limit gives:
\[ = \frac{1}{b} \lim_{u \to 0} \frac{\sin(u)}{\frac{u}{a}} \cdot \sin(u) \]

### Step 5: Simplify the limit
Now we can simplify:
\[ = \frac{1}{b} \cdot a \cdot \lim_{u \to 0} \sin(u) = \frac{a}{b} \cdot 0 = 0 \]

### Conclusion
Thus, the final result is:
\[ \lim_{x \to 0} \frac{\sin^2(ax)}{bx} = 0 \]