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

Answer

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

### Step 1: Substitute \( x = 0 \)

First, we substitute \( x = 0 \) into the limit expression:

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

Now we evaluate the denominator:

\[ \sec(0) = \frac{1}{\cos(0)} = 1 \]

Thus,

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

So the limit becomes:

\[ \frac{0}{\sqrt{2}} = 0 \]

### Step 2: Analyze the limit

Since we have a \( \frac{0}{0} \) form, we can apply L'Hôpital's Rule. However, in this case, we can simplify the expression further before applying L'Hôpital's Rule.

### Step 3: Rewrite the limit

We can rewrite the limit as follows:

\[ \lim_{x \to 0} \frac{\sin^2(3x)}{\sqrt{3 + \sec x - 2}} = \lim_{x \to 0} \frac{\sin^2(3x)}{\sqrt{1 + \sec x - 1}} = \lim_{x \to 0} \frac{\sin^2(3x)}{\sqrt{\sec x - 1}} \]

### Step 4: Use Taylor expansion for small \( x \)

For small \( x \), we can use the Taylor series expansion:

\[ \sec x \approx 1 + \frac{x^2}{2} + O(x^4) \]

Thus,

\[ \sec x - 1 \approx \frac{x^2}{2} \]

So we have:

\[ \sqrt{\sec x - 1} \approx \sqrt{\frac{x^2}{2}} = \frac{|x|}{\sqrt{2}} \]

### Step 5: Substitute back into the limit

Now substituting this back into our limit, we have:

\[ \lim_{x \to 0} \frac{\sin^2(3x)}{\frac{|x|}{\sqrt{2}}} \]

Using \( \sin(3x) \approx 3x \) for small \( x \):

\[ \sin^2(3x) \approx (3x)^2 = 9x^2 \]

Thus, the limit becomes:

\[ \lim_{x \to 0} \frac{9x^2}{\frac{|x|}{\sqrt{2}}} = \lim_{x \to 0} \frac{9\sqrt{2}x^2}{|x|} = \lim_{x \to 0} 9\sqrt{2} |x| = 0 \]

### Conclusion

Therefore, the value of the limit is:

\[ \boxed{0} \]