The value of `lim_(xrarr0)(sin^(2)3x)/(sqrt(3+secx-2))` is equal to

Answer

To solve the limit problem \( \lim_{x \to 0} \frac{\sin^2(3x)}{\sqrt{3 + \sec x - 2}} \), we will follow these steps:

### Step 1: Substitute \( x = 0 \)
First, we substitute \( x = 0 \) into the limit expression to check if it results in an indeterminate form.

\[ \sin^2(3 \cdot 0) = \sin^2(0) = 0 \]

Next, we evaluate the denominator:

\[ \sec(0) = \frac{1}{\cos(0)} = 1 \]
\[ \sqrt{3 + \sec(0) - 2} = \sqrt{3 + 1 - 2} = \sqrt{2} \]

So, we have:

\[ \frac{\sin^2(3x)}{\sqrt{3 + \sec x - 2}} \bigg|_{x=0} = \frac{0}{\sqrt{2}} = 0 \]

### Step 2: Conclusion
Since substituting \( x = 0 \) gives us \( \frac{0}{\sqrt{2}} \), which is not an indeterminate form, we conclude that:

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

Thus, the final answer is:

\[ \boxed{0} \]