Question

limita cand x tinde la 3 din ( radical2x+3-radical x+6)/(radical 4x+4-radical 5x+1)

Answer 1

Mai intai trebuie sa ne ocupam de rationalizarea radicalilor:
$L= \lim_{x \to 3} \frac{ \sqrt{2x+3} - \sqrt{x+6} }{\sqrt{4x+4}-\sqrt{5x+1}} = \lim_{x \to 3} \frac{\frac{2x+3-x-6}{\sqrt{2x+3} + \sqrt{x+6}}}{\frac{4x+4-5x-1}{\sqrt{4x+4}+\sqrt{5x+1}}}$
De aici, facand calcule si simplificari, obtinem:
[tex]L= \lim_{x\to3}\frac{(x-3)(\sqrt{4x+4}+\sqrt{5x+1})}{(-x+3)(\sqrt{2x+3} + \sqrt{x+6})} =-\lim_{x\to3}\frac{\sqrt{4x+4}+\sqrt{5x+1}}{\sqrt{2x+3} + \sqrt{x+6}}[/tex]
Am simplificat x-3 si -x+3, deci a ramas un -1 pe care l-am scos in fata limitei. De aici, doar inlocuim 3 in forma care ne-a ramas si obtinem limita:
$L= -\frac{\sqrt{4*3+4}+\sqrt{5*3+1}}{\sqrt{2*3+3} + \sqrt{3+6}} = -\frac{4+4}{3+3}=-\frac{8}{6}=-\frac{4}{3}$