Evaluate the following limits :
`Lim_(x to 0) (2 sin^(2) 3x)/(x^(2))`

Answer

To evaluate 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 identity for sine
We know that \( \sin(\theta) \approx \theta \) as \( \theta \) approaches 0. Thus, we can use the identity:
\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]
In our case, let \( u = 3x \). Then, as \( x \to 0 \), \( u \to 0 \) as well.

### Step 3: Modify the limit
We can express \( \sin^2(3x) \) in terms of \( u \):
\[ \sin^2(3x) = \sin^2(u) \]
Now we rewrite our limit:
\[ \lim_{x \to 0} \frac{2 \sin^2(3x)}{x^2} = \lim_{x \to 0} \frac{2 \sin^2(u)}{(u/3)^2} \cdot \frac{(u/3)^2}{x^2} \]
Since \( u = 3x \), we have \( x = \frac{u}{3} \), thus \( x^2 = \frac{u^2}{9} \).

### Step 4: Substitute and simplify
Substituting \( x^2 \) into the limit gives:
\[ \lim_{u \to 0} \frac{2 \sin^2(u)}{(u/3)^2} = \lim_{u \to 0} \frac{2 \sin^2(u)}{\frac{u^2}{9}} = \lim_{u \to 0} \frac{18 \sin^2(u)}{u^2} \]

### Step 5: Apply the limit identity
Now we can apply the limit identity:
\[ \lim_{u \to 0} \frac{\sin^2(u)}{u^2} = 1 \]
Thus,
\[ \lim_{u \to 0} \frac{18 \sin^2(u)}{u^2} = 18 \cdot 1 = 18 \]

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

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