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

To evaluate the limit \( \lim _{x \rightarrow 3} \frac{\sqrt{x+1}-2}{x-3} \), we first substitute \( x = 3 \). This substitution leads us to an indeterminate form \( \frac{0}{0} \). To resolve this, we can manipulate the expression by multiplying the numerator and denominator by the conjugate of the numerator: \[ \frac{\sqrt{x+1}-2}{x-3} \cdot \frac{\sqrt{x+1}+2}{\sqrt{x+1}+2} = \frac{(\sqrt{x+1}-2)(\sqrt{x+1}+2)}{(x-3)(\sqrt{x+1}+2)} = \frac{(x+1)-4}{(x-3)(\sqrt{x+1}+2)} = \frac{x-3}{(x-3)(\sqrt{x+1}+2)}. \] Now we can cancel \( x-3 \) (since \( x \neq 3 \) in the limit): \[ = \frac{1}{\sqrt{x+1}+2}. \] Now, we evaluate the limit as \( x \) approaches 3: \[ \lim_{x \rightarrow 3} \frac{1}{\sqrt{x+1}+2} = \frac{1}{\sqrt{3+1}+2} = \frac{1}{\sqrt{4}+2} = \frac{1}{2+2} = \frac{1}{4}. \] Thus, the limit is: \[ \lim _{x \rightarrow 3} \frac{\sqrt{x+1}-2}{x-3} = \frac{1}{4}. \]