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

To evaluate the limit \( \lim _{x \rightarrow 2} \frac{\sqrt{x+2}-2}{\sqrt{2 x-3}-1} \), we start by substituting \( x = 2 \): \[ \frac{\sqrt{2+2}-2}{\sqrt{2 \cdot 2-3}-1} = \frac{\sqrt{4}-2}{\sqrt{4-3}-1} = \frac{2-2}{1-1} = \frac{0}{0} \] This indeterminate form necessitates further analysis, often involving algebraic manipulation. We can rationalize the numerator and the denominator to simplify our limit expression. ### Step 1: Rationalize the Numerator Multiply the numerator and denominator by the conjugate of the numerator: \[ \frac{\sqrt{x+2}-2}{\sqrt{2x-3}-1} \cdot \frac{\sqrt{x+2}+2}{\sqrt{x+2}+2} \] This will give: \[ \frac{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}{(\sqrt{2x-3}-1)(\sqrt{x+2}+2)} = \frac{x+2-4}{(\sqrt{2x-3}-1)(\sqrt{x+2}+2)} = \frac{x-2}{(\sqrt{2x-3}-1)(\sqrt{x+2}+2)} \] ### Step 2: Rationalize the Denominator for the New Expression Now look at the remaining denominator \( \sqrt{2x-3}-1 \) which also approaches 0 as \( x \) approaches 2. We can rationalize this by multiplying the denominator by \( \sqrt{2x-3}+1 \): \[ \frac{x-2}{(\sqrt{2x-3}-1)(\sqrt{x+2}+2)} \cdot \frac{\sqrt{2x-3}+1}{\sqrt{2x-3}+1} \] This results in: \[ \frac{(x-2)(\sqrt{2x-3}+1)}{(2x-3-1)(\sqrt{x+2}+2)} = \frac{(x-2)(\sqrt{2x-3}+1)}{(2x-4)(\sqrt{x+2}+2)} = \frac{(x-2)(\sqrt{2x-3}+1)}{2(x-2)(\sqrt{x+2}+2)} \] ### Step 3: Cancel Common Terms Assuming \( x \neq 2 \), we can safely cancel \( (x-2) \): \[ \frac{\sqrt{2x-3}+1}{2(\sqrt{x+2}+2)} \] ### Step 4: Compute the Limit Now we can compute the limit as \( x \) approaches 2: \[ \lim _{x \rightarrow 2} \frac{\sqrt{2x-3}+1}{2(\sqrt{x+2}+2)} = \frac{\sqrt{2(2)-3}+1}{2(\sqrt{2+2}+2)} = \frac{\sqrt{1}+1}{2(\sqrt{4}+2)} = \frac{1+1}{2(2+2)} = \frac{2}{8} = \frac{1}{4} \] Thus, the limit is: \[ \boxed{\frac{1}{4}} \]