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

Answer

To evaluate the limit

\[ \lim_{x \to 0} \frac{\tan(3x) - 2x}{3x - \sin^2(x)}, \]

we will follow these steps:

### Step 1: Rewrite the limit expression
We start by rewriting the limit expression by dividing the numerator and the denominator by \(x\):

\[ \lim_{x \to 0} \frac{\tan(3x) - 2x}{3x - \sin^2(x)} = \lim_{x \to 0} \frac{\frac{\tan(3x)}{x} - 2}{3 - \frac{\sin^2(x)}{x}}. \]

### Step 2: Simplify the limit
Now we simplify the expression:

\[ = \lim_{x \to 0} \left( \frac{\tan(3x)}{x} - 2 \right) \Big/ \left( 3 - \frac{\sin^2(x)}{x} \right). \]

### Step 3: Apply L'Hôpital's Rule
Since both the numerator and denominator approach 0 as \(x\) approaches 0, we can apply L'Hôpital's Rule, which states that if the limit results in an indeterminate form \( \frac{0}{0} \), we can differentiate the numerator and denominator:

1. Differentiate the numerator: The derivative of \(\tan(3x)\) is \(3 \sec^2(3x)\).
2. Differentiate the denominator: The derivative of \(3x\) is \(3\) and the derivative of \(\sin^2(x)\) can be found using the chain rule: \(2\sin(x)\cos(x) = \sin(2x)\).

Thus, we have:

\[ = \lim_{x \to 0} \frac{3 \sec^2(3x)}{3 - \sin(2x)}. \]

### Step 4: Evaluate the limit
Now we can substitute \(x = 0\):

\[ = \frac{3 \sec^2(0)}{3 - \sin(0)} = \frac{3 \cdot 1}{3 - 0} = \frac{3}{3} = 1. \]

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

\[ \lim_{x \to 0} \frac{\tan(3x) - 2x}{3x - \sin^2(x)} = 1. \]