4) \( \lim _{x \rightarrow 1} \frac{\sqrt{x+3}-\sqrt{4 x}}{x^{2}+x-2} \)

To solve the limit \( \lim _{x \rightarrow 1} \frac{\sqrt{x+3}-\sqrt{4 x}}{x^{2}+x-2} \), we first evaluate the expression at \( x = 1 \): Numerator: \[ \sqrt{1+3} - \sqrt{4 \cdot 1} = \sqrt{4} - \sqrt{4} = 2 - 2 = 0 \] Denominator: \[ 1^2 + 1 - 2 = 1 + 1 - 2 = 0 \] We have an indeterminate form \( \frac{0}{0} \), so we need to simplify the expression. Let's factor the denominator \( x^2 + x - 2 \). It can be factored as: \[ x^2 + x - 2 = (x - 1)(x + 2) \] Now we rewrite the limit: \[ \lim_{x \rightarrow 1} \frac{\sqrt{x+3}-\sqrt{4x}}{(x-1)(x+2)} \] Next, we can simplify the numerator by multiplying the numerator and denominator by the conjugate of the numerator \( \sqrt{x+3} + \sqrt{4x} \): \[ \lim_{x \rightarrow 1} \frac{(\sqrt{x+3}-\sqrt{4x})(\sqrt{x+3}+\sqrt{4x})}{(x-1)(x+2)(\sqrt{x+3}+\sqrt{4x})} \] Simplifying further: \[ \sqrt{x+3}^2 - \sqrt{4x}^2 = (x + 3) - (4x) = -3x + 3 = 3(1 - x) \] Now our limit expression becomes: \[ \lim_{x \rightarrow 1} \frac{3(1 - x)}{(x - 1)(x + 2)(\sqrt{x + 3} + \sqrt{4x})} \] Notice that \( 1 - x = -(x - 1) \): \[ = \lim_{x \rightarrow 1} \frac{-3(x - 1)}{(x - 1)(x + 2)(\sqrt{x + 3} + \sqrt{4x})} \] We can cancel out \( (x - 1) \) (as long as \( x \neq 1 \)): \[ = \lim_{x \rightarrow 1} \frac{-3}{(x + 2)(\sqrt{x + 3} + \sqrt{4x})} \] Now substitute \( x = 1 \) into the expression: \[ = \frac{-3}{(1 + 2)(\sqrt{1 + 3} + \sqrt{4 \cdot 1})} = \frac{-3}{3(2)} = \frac{-3}{6} = -\frac{1}{2} \] Thus, the limit is: \[ \boxed{-\frac{1}{2}} \]