Question

URGENT!!CU HOSPITAL VA ROG!!!

Answer 1

Inlocuind pe x cu 2 in limita obtinem cazul de nedeterminare $\frac{0}{0}$.
Deci putem aplica L'Hopital: derivam numaratorul si numitorul.

$\lim_{x\to 2} \cfrac{(\sqrt{x^2+5}-\sqrt{x^2+4x-3})'}{(x^2-2x)'}=$

$=\lim_{x\to 2}\cfrac{\cfrac{1}{2\sqrt{x^2+5}}(x^2+5)'-\cfrac{1}{2\sqrt{x^2+4x-3}}(x^2+4x-3)'}{2x-2}}=$

$=\lim_{x\to 2}\cfrac{\cfrac{2x}{2\sqrt{x^2+5}}-\cfrac{2x+4}{2\sqrt{x^2+4x+3}}}{2x-2}=$

$=\cfrac{\cfrac{4}{2\times3}-\cfrac{4+4}{2\times3}}{4-2}=\cfrac{4-8}{2\times6}=\cfrac{-4}{12}=-\cfrac{1}{3}$

Answer 2

Răspuns:

-1/3

Explicație pas cu pas:

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