`lim_(x to 0) (2 sin^(2) 3x)/x^(2)` =

Options

A. 9
B. 2
C. 18
D. 3

Answer

To solve the limit \( \lim_{x \to 0} \frac{2 \sin^2(3x)}{x^2} \), we can follow these steps:

### Step 1: Rewrite the limit
We start with the limit:
\[ \lim_{x \to 0} \frac{2 \sin^2(3x)}{x^2} \]

### Step 2: Use the substitution
We can substitute \( u = 3x \). Then as \( x \to 0 \), \( u \to 0 \) as well. We also need to express \( x \) in terms of \( u \):
\[ x = \frac{u}{3} \]
Thus, \( x^2 = \left(\frac{u}{3}\right)^2 = \frac{u^2}{9} \).

### Step 3: Substitute into the limit
Now, substituting \( u \) into the limit gives:
\[ \lim_{u \to 0} \frac{2 \sin^2(u)}{\left(\frac{u}{3}\right)^2} = \lim_{u \to 0} \frac{2 \sin^2(u)}{\frac{u^2}{9}} = \lim_{u \to 0} \frac{2 \sin^2(u) \cdot 9}{u^2} = 18 \lim_{u \to 0} \frac{\sin^2(u)}{u^2} \]

### Step 4: Use the limit property
We know from the standard limit property that:
\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]
Thus, we can square this limit:
\[ \lim_{u \to 0} \frac{\sin^2(u)}{u^2} = \left(\lim_{u \to 0} \frac{\sin(u)}{u}\right)^2 = 1^2 = 1 \]

### Step 5: Combine the results
Now substituting back into our limit:
\[ 18 \lim_{u \to 0} \frac{\sin^2(u)}{u^2} = 18 \cdot 1 = 18 \]

### Final Answer
Thus, the limit is:
\[ \lim_{x \to 0} \frac{2 \sin^2(3x)}{x^2} = 18 \]

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