`lim_(x to oo ) (sqrt(3x^(2)-1)-sqrt(2x^(2)-3))/(4x+3)`

Answer

To solve the limit problem \( \lim_{x \to \infty} \frac{\sqrt{3x^2 - 1} - \sqrt{2x^2 - 3}}{4x + 3} \), we can follow these steps:

### Step 1: Factor out \( x \) from the square roots in the numerator
We start by rewriting the square roots in the numerator:

\[ \sqrt{3x^2 - 1} = \sqrt{x^2(3 - \frac{1}{x^2})} = x\sqrt{3 - \frac{1}{x^2}} \]

\[ \sqrt{2x^2 - 3} = \sqrt{x^2(2 - \frac{3}{x^2})} = x\sqrt{2 - \frac{3}{x^2}} \]

Thus, we can rewrite the limit as:

\[ \lim_{x \to \infty} \frac{x\sqrt{3 - \frac{1}{x^2}} - x\sqrt{2 - \frac{3}{x^2}}}{4x + 3} \]

### Step 2: Simplify the expression
Now, we can factor \( x \) out of the numerator:

\[ = \lim_{x \to \infty} \frac{x(\sqrt{3 - \frac{1}{x^2}} - \sqrt{2 - \frac{3}{x^2}})}{4x + 3} \]

### Step 3: Factor out \( x \) from the denominator
We can also factor \( x \) out of the denominator:

\[ = \lim_{x \to \infty} \frac{x(\sqrt{3 - \frac{1}{x^2}} - \sqrt{2 - \frac{3}{x^2}})}{x(4 + \frac{3}{x})} \]

### Step 4: Cancel \( x \) from numerator and denominator
Now we can cancel \( x \) (assuming \( x \neq 0 \)):

\[ = \lim_{x \to \infty} \frac{\sqrt{3 - \frac{1}{x^2}} - \sqrt{2 - \frac{3}{x^2}}}{4 + \frac{3}{x}} \]

### Step 5: Evaluate the limit as \( x \to \infty \)
As \( x \) approaches infinity, the terms \( \frac{1}{x^2} \) and \( \frac{3}{x} \) approach 0:

\[ = \frac{\sqrt{3 - 0} - \sqrt{2 - 0}}{4 + 0} = \frac{\sqrt{3} - \sqrt{2}}{4} \]

### Final Answer
Thus, the limit is:

\[ \lim_{x \to \infty} \frac{\sqrt{3x^2 - 1} - \sqrt{2x^2 - 3}}{4x + 3} = \frac{\sqrt{3} - \sqrt{2}}{4} \]

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