\lim\limits_{x \to 4 }\frac{\sqrt{2x-7}-1}{\sqrt{x-3}-1}=\frac{\sqrt{2\cdot 4-7}-1}{\sqrt{4-3}-1}=\frac{1-1}{1-1}=\frac{0}{0} wyrażenie nieoznaczone
Liczę inaczej
a-b=\frac{a^2-b^2}{a+b}
{\lim\limits_{x \to 4} \frac{\sqrt{2x-7}-1}{\sqrt{x-3}-1}=\frac{\frac{(\sqrt{2x-7})^2-1^2}{\sqrt{2x-7}+1}}{\frac{(\sqrt{x-3})^2-1^2}{\sqrt{x-3}+1}}=\frac{\frac{2x-7-1}{2x-7+1}}{\frac{x-3-1}{x-3+1}}=\frac{\frac{2x-8}{2x-6}}{\frac{x-4}{x-2}}=\frac{2(x-4)}{2(x-3)}\cdot \frac{x-2}{x-3}=\frac{x-2}{x-3}=\frac{4-2}{4-3}=2}