Evaluate the following limits:
`lim_(xrarr0)((1-cos3x))/(x^(2))`

Explanation

To evaluate the limit \( \lim_{x \to 0} \frac{1 - \cos(3x)}{x^2} \), we can use the trigonometric identity and properties of limits. Here’s a step-by-step solution:

### Step 1: Use the identity for \( 1 - \cos(θ) \)
We know from trigonometric identities that:
\[ 1 - \cos(θ) = 2 \sin^2\left(\frac{θ}{2}\right) \]
Applying this identity with \( θ = 3x \):
\[ 1 - \cos(3x) = 2 \sin^2\left(\frac{3x}{2}\right) \]
Thus, we can rewrite our limit as:
\[ \lim_{x \to 0} \frac{1 - \cos(3x)}{x^2} = \lim_{x \to 0} \frac{2 \sin^2\left(\frac{3x}{2}\right)}{x^2} \]

### Step 2: Rewrite the limit
Now, we can factor out the constant:
\[ = 2 \lim_{x \to 0} \frac{\sin^2\left(\frac{3x}{2}\right)}{x^2} \]

### Step 3: Change the variable
To simplify further, we can change the variable. Let \( u = \frac{3x}{2} \). Then, as \( x \to 0 \), \( u \to 0 \) as well. Notice that \( x = \frac{2u}{3} \), so \( x^2 = \left(\frac{2u}{3}\right)^2 = \frac{4u^2}{9} \). Substituting this into the limit gives:
\[ = 2 \lim_{u \to 0} \frac{\sin^2(u)}{\left(\frac{4u^2}{9}\right)} = 2 \lim_{u \to 0} \frac{9 \sin^2(u)}{4u^2} \]

### Step 4: Apply the limit property
We know that:
\[ \lim_{u \to 0} \frac{\sin(u)}{u} = 1 \]
Thus:
\[ \lim_{u \to 0} \frac{\sin^2(u)}{u^2} = \left(\lim_{u \to 0} \frac{\sin(u)}{u}\right)^2 = 1^2 = 1 \]
So we can simplify our limit:
\[ = 2 \cdot \frac{9}{4} \cdot 1 = \frac{18}{4} = \frac{9}{2} \]

### Final Answer
Therefore, the limit is:
\[ \lim_{x \to 0} \frac{1 - \cos(3x)}{x^2} = \frac{9}{2} \]