Since we have ambiguity of 0/0, we can use l'Hopital rule: <img src="/cgi-bin/mimetex.cgi?%5Clim_%7Bx%20%5Cto%201%7D%20%5Cfrac%7B%28%5Csqrt%7B5x%7D%20-5%29%27%7D%7B%28%5Csqrt%7B2x-4%7D-%20%5Csqrt%7B6%7D%29%27%7D%20=%20%5Clim_%7Bx%20%5Cto%201%7D%20%5Cfrac%7B5/%282%5Csqrt%7B5x%7D%29%7D%7B2/%282%5Csqrt%7B2x-4%7D%29%7D%20=%20%5Clim_%7Bx%20%5Cto%201%7D%20%5Cfrac%7B5%20%5Csqrt%7B2x-4%7D%7D%7B2%5Csqrt%7B5x%7D%7D%20=%20%5Cfrac%7B%5Csqrt%7B6%7D%7D%7B2%7D" title="\lim_{x \to 1} \frac{(\sqrt{5x} -5)'}{(\sqrt{2x-4}- \sqrt{6})'} = \lim_{x \to 1} \frac{5/(2\sqrt{5x})}{2/(2\sqrt{2x-4})} = \lim_{x \to 1} \frac{5 \sqrt{2x-4}}{2\sqrt{5x}} = \frac{\sqrt{6}}{2}">
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